MATH 2501 Notes

Multivariable Calculus and Linear Algebra

As taught by Prof. Bruce Hughes
Vanderbilt University - Spring Semester 2016
Transcribed by David K. Zhang


The notes below were taken live in-class during Prof. Hughes' MATH 2501 lectures at Vanderbilt University. They are provided here exclusively as a historical record of what topics were covered during each day of class. For readers who aim to learn this material for the first time, they are incomplete in several crucial ways:

  • Gaps in the logical flow of ideas due to missed classes.
  • Omitted discussions, proofs, and examples which I considered obvious.
  • Missing sections which were written down on paper and never typeset.

Time permitting, I plan to release an expanded and revised version of these notes in the future to address these deficiencies.

Lecture 01 (2016-01-11)

Introduction to manifolds. The main new topic of the second semester will be integration of functions $f: \R^n \to \R$. We will generalize the one-dimensional integral

\[\int_a^b f(x) \dd{x} = \int_{[a,b]} f(x) \dd{x} \]

to more exciting higher-dimensional domains called manifolds (possibly with boundary).

Rough idea of manifolds. We call $M \subseteq \R^n$ a $k$-dimensional manifold if each point in $M$ has a neighborhood that can be described by a $k$-dimensional curvilinear coordinate system.

Definition: $M \subseteq \R^n$ is a $k$-dimensional manifold if for each $\vp \in M$, there exist open sets $U \subseteq \R^k$ and $W \subseteq \R^n$ and a function $g: U \to \R^n$ such that

  1. $\vp \in W$.

  2. $g(U) = M \cap W$.

  3. $g$ is smooth (i.e., infinitely differentiable).

  4. $g$ in injective.

  5. $g^{-1}: M \cap W \to U$ is continuous.

  6. For all $\vx \in U$, $\rank(Jg(\vx)) = k$. (Recall that $\rank(Jg(\vx))$ is the dimension of the image of $Dg(\vx)$. We call this image the tangent space to the image of $g$ at $g(\vx)$.)

Example: Let $M \subseteq \R^3$ be the solution set of $z = x^2 + y^2$. This is the unit paraboloid, i.e., the graph of $f(x,y) = x^2 + y^2$. Given $\vp \in M$, let $U = \R^2$ and $W = \R^3$. Define

\[g(s,t) = \mqty[s\\t\\s^2+t^2] \]

Then, we have that

  1. $\vp \in W$.

  2. $g(U) = M = M \cap W$.

  3. $g$ is smooth (since it has polynomial components).

  4. $g$ is injective.

  5. $g^{-1}: M \to \R^2$ is given by $g^{-1}(x,y,z) = (x,y)$.

  6. $Jg(s,t) = \mqty[1&0\\0&1\\2s&2t]$ has rank 2.

Hence, $M$ is a 2-dimensional manifold.

Lecture 02 (2016-01-13)

Recall: $M \subseteq \R^n$ is a $k$-dimensional manifold if $\forall \vp \in M$ $\exists U \opensubset \R^k$, $W \opensubset \R^n$, $g: U \bijto M \cap W$ such that

  1. $\vp \in W$.

  2. $g$ is smooth.

  3. $g^{-1}$ is continuous.

  4. $\forall \vx \in U,\ \rank Jg(\vx) = k$.

Manifold Characterization Theorem: Let $M \subseteq \R^n$. The following are equivalent:

  1. $M$ is a $k$-dimensional manifold in $\R^n$.

  2. $\forall \vp \in M$ $\exists W \opensubset \R^n$, $F: W \to \R^{n-k}$ such that

  3. $\vp \in W$.

  4. $M \cap W = F^{-1}(\vo)$.

  5. $F$ is smooth.

  6. $\forall \vx \in M$, $\rank JF(\vx) = n-k$.

Remark: This theorem gives an implicit characterization of manifolds as local level sets of smooth locally invertible functions. Its proof requires machinery which will be developed later in the course.


  1. The sphere $S^{n-1} \subseteq \R^n$ is defined by

    \[S^{n-1} = \{ \vx \in \R^n : \norm{\vx} = 1 \}. \]

    Define $F: \R^n \to \R$ by $F(\vx) = x_1^2 + \dots + x_n^2 - 1$, and let $W = \R^n$. Then $S^{n-1} = F^{-1}(0)$, $F$ is smooth, and

    \[JF(\vx) = \mqty[2x_1 & 2x_2 & \cdots & 2x_n] \]

    has rank $1$ for all $\vx \in S^{n-1}$. Hence, by the manifold characterization theorem, $S^{n-1}$ is a manifold.

    We can also verify that $S^{n-1}$ is a manifold directly by definition. (This part omitted. Proceeds straightforwardly by pushing a disk up to a half-sphere.)

  2. Consider the configuration space $M$ of all ordered pairs of orthogonal unit vectors in $\R^3$.

    \[M = \{ (\vva, \vvb) \in \R^3 \times \R^3 : \norm{\vva} = \norm{\vvb} = 1,\ \vva \cdot \vvb = 0 \}. \]

    Let $W = \R^6$ and define $F: \R^6 \to \R^3$ by

    \[F(x_1, \dots, x_6) = \mqty[x_1^2 + x_2^2 + x_3^2 - 1 \\ x_4^2 + x_5^2 + x_6^2 - 1 \\ x_1x_4 + x_2x_5 + x_3x_6]. \]

    Then $M = F^{-1}(\vo)$, $F$ is smooth, and

    \[JF(\vx) = \mqty[2x_1 & 2x_2 & 2x_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2x_4 & 2x_5 & 2x_6 \\ x_4 & x_5 & x_6 & x_1 & x_2 & x_3] \]

    has rank $3$ (any smaller and $\vva$ would be a scalar multiple of $\vvb$).

Lecture 03 (2016-01-15)

The Inverse Function Theorem. Let $\vp \in U \opensubset \R^n$, and suppose $f: U \to \R^n$ is a smooth function with $[Df(\vp)] = Jf(\vp)$ invertible. Then $f$ is invertible near $f(\vp)$. More precisely, there exists $V \opensubset \R^n$ such that $\vp \in V \subseteq U$, $f|_V : V \to f(V)$ is a bijection, and the inverse $g: f(V) \to V$ of $f|_V$ is smooth. Moreover, $Jg(f\vp) = Jf(\vp)^{-1}$.

Example: Let $f: \R^2 \to \R^2$ be defined by $f(x,y) = (x^2,y)$. Then near $\vp = (2,1)$, a local inverse of $f$ is given by $g(s,t) = (\sqrt{s}, t)$. In this case, we can take

\[V = f(V) = \{ (x,y) \in \R^2 : x > 0 \} \]

and for $(x,y) \in V$, we have

\[Jf(x,y) = \mqty[2x&0\\0&1] \qquad Jg(f(x,y)) = Jg(x^2,y) = \mqty[\frac{1}{2x}&0\\0&1] \]

so indeed $Jg(f(x,y)) = Jf(x,y)^{-1}$.

Using IFT to prove the Manifold Characterization Theorem. We will show that (II) implies (I). (The converse will be proven later.) Recall that we are given a smooth function $F: \R^n \to \R^{n-k}$ with $\rank JF(\vx) = n-k$ for all $\vx \in M \coloneqq F^{-1}(\vo)$. We want to show that $M$ is a $k$-dimensional manifold.

Let $\vp \in M$ be given. Since $\rank JF(\vp) = n-k$, there are $k$ free variables in the system $JF(\vp)\vx = \vo$. For notational simplicity, assume the free variables are $x_1, \dots, x_k$. Then the RREF of $JF(\vp)$ is $\mqty[0 & I_{n-k}]$. Define $f: \R^n \to \R^n$ by

\[f(\vx) = \mqty[x_1\\\vdots\\x_k\\F_1(\vx)\\\vdots\\F_{n-k}(\vx)]. \]


\[Jf(\vp) = \mqty[I_k & 0 \\ \multicolumn{2}{c}{JF(\vp)}] \rightsquigarrow \mqty[I_k & 0 \\ 0 & I_{n-k}] = I_n, \]

Hence, $Jf(\vp)$ is invertible, and IFT applies, giving us $\vp \in V \opensubset \R^n$ with $f(V) \opensubset \R^n$ where $f|_V: V \to f(V)$ has a smooth inverse $g: f(V) \to V$. This gives the necessary data to show that $M$ is a $k$-dimensional manifold. QED

Lecture 04 (2016-01-20)

Banach's Contraction Mapping Principle. This will be our main tool for proving the implicit function theorem.

Definition: Let $X \subseteq \R^n$. A function $f: X \to X$ is a contraction mapping if $\exists c \in \R$ such that $0 \le c < 1$ and

\[\norm{f(\vx) - f(\vy)} \le c\norm{\vx - \vy} \]

for all $\vx, \vy \in X$.


  1. Let $f: \R^n \to \R^n$ be defined by $f(\vx) = \frac{1}{4} \vx + \ve_1$. Then $\norm{f(\vx) - f(\vy)} = \frac{1}{4} \norm{\vx - \vy}$, so $f$ is a contraction mapping with $c = \frac{1}{4}$.

  2. Let $f: \R^n \to \R^n$ be defined by $f(\vx) = \vx + 5\ve_1$. Then $\norm{f(\vx) - f(\vy)} = \norm{\vx - \vy}$, so $f$ is not a contraction mapping.

Theorem (CMP): Suppose $X \subseteq \R^n$ is nonempty and closed. If $f: X \to X$ is a contraction mapping, then there exists a unique $\vx \in X$ such that $f(\vx) = \vx$. (In other words, $f$ has a unique fixed point.)


  1. Take $f: \R^n \to \R^n$ from example (1) above. Then the unique fixed point of $f$ is $\vx = \frac{4}{3} \ve_1$.

  2. Every linear transformation $T: \R^n \to \R^n$ has a fixed point $\vo$. This may or may not be the only fixed point.

Two Facts about Infinite Series.

  1. If $a_k \in \R$, $a_k \ge 0$ for all $k = 1, 2, 3, \dots$, and $\{s_n\}_{n=1}^\infty$ is bounded, then $\sum_{k=1}^\infty a_k$ converges.

  2. If $\sum_{k=1}^\infty \norm{\vva_k}$ converges, then so does $\sum_{k=1}^\infty \vva_k$.

Proof of Existence in CMP: Fix any $\vx_0 \in X$, and let $\vx_{k+1} = f(\vx_k)$. We claim that $\{\vx_k\}_{k=1}^\infty$ converges. To see this, let $\vva_k = \vx_k - \vx_{k-1}$. Then

\[\norm{\vva_k} = \norm{\vx_k - \vx_{k-1}} = \norm{f(\vx_{k-1}) - f(\vx_{k-2})} \le c\norm{\vx_{k-1} - \vx_{k-2}} = c\norm{\vva_{k-1}}. \]

This shows that $\{\norm{\vva_k}\}_{k=1}^\infty$ is a positive series bounded above by a geometric series. It follows by fact 1 that $\sum_{k=1}^\infty \norm{\vva_k}$ converges, and by fact 2 that $\sum_{k=1}^\infty \vva_k$ converges. Since $X$ is closed, $\vx \coloneqq \lim_{k \to \infty} \vx_k = \vx_0 + \vva$ exists and is a member of $X$.

By homework problem #3, every contraction mapping is continuous. It follows that

\[\vx = \lim_{k \to \infty} \vx_k = \lim_{k \to \infty} \vx_{k+1} = \lim_{k \to \infty} f(\vx_k) = f(\vx), \]

as desired. QED

(Proving uniqueness is a homework problem.)

Lecture 4.5 (2016-01-21)

The Operator Norm and Mean Value Theorem. These will be useful tools for the proof of the inverse function theorem.

Example: We illustrate how the univariate MVT can be used to show that a function is a contraction mapping. Define $f: [0,1] \to [0,1]$ by $f(x) = \cos x$. By the mean value theorem, whenever $0 \le x < y \le 1$, there exists $z \in [x,y]$ such that

\[f(x) - f(y) = f'(z)(x-y). \]

It follows that

\[\abs{f(x) - f(y)} = \abs{f'(z)} \abs{(x-y)} \]

and since $f'(x) = \sin x$ is bounded above on $[0,1]$ by $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} < 1$, we have that $f$ is a contraction mapping, since

\[\abs{f(x) - f(y)} \le \frac{\sqrt{3}}{2} \abs{(x-y)}. \]

Moreover, if we wanted to, we could numerically solve the equation $x = \cos x$ by taking any $x_0 \in [0,1]$ and repeatedly taking its cosine.

Definition: Let $T: \R^n \to \R^m$ be a linear transformation. The operator norm of $T$ is

\[\norm{T} \coloneqq \sup_{\norm{\vx} = 1} \norm{T(\vx)}. \]

Note that this sup exists by the maximum value theorem.

Theorem: For all $\vx \in \R^n$, we have $\norm{T(\vx)} \le \norm{T} \norm{\vx}$.

Proof: The result holds trivially for $\vx = \vo$. For nonzero $\vx$, we have

\[\frac{1}{\norm{\vx}} \norm{T(\vx)} = \norm{T\qty(\frac{\vx}{\norm{\vx}})} \le \norm{T} \]

by definition. Hence,

\[\norm{T(\vx)} \le \norm{T} \norm{\vx}, \]

as desired. QED

Corollary: If $T: \R^n \to \R^n$ is a linear transformation such that $\norm{T} < 1$, then $T$ is a contraction mapping.

Proof: Let $c = \norm{T}$. Then for all $\vx, \vy \in \R^n$, we have

\[\norm{T(\vx) - T(\vy)} = \norm{T(\vx - \vy)} \le \norm{T} \norm{\vx - \vy} = c\norm{\vx - \vy}. \]


Remark: Note that any such $T$ has $\vo$ as its unique fixed point.

Example: Let $T: \R^n \to \R$ be a linear transformation, and let $\vv = [T]^\top$. Then $T(\vx) = \vv \cdot \vx$ for all $\vx \in \R^n$, and $\norm{T} = \norm{\vv}$. To see this, observe by Cauchy-Schwarz that

\[\norm{T(\vx)} = \norm{\vv \cdot \vx} \le \norm{\vv} \norm{\vx} \]

This means that $\norm{T}$ is bounded above by $\norm{\vv}$. Moreover, if $\vv \ne \vo$, then

\[\norm{\vv} = \norm{\frac{\vv \cdot \vv}{\norm{\vv}}} = \norm{T\qty(\frac{\vv}{\norm{\vv}})} \le \norm{T} \]

giving a bound below. For the $\vv = \vo$ case, we have $\norm{T} = \norm{\vv} = 0$.

Multivariable Mean Value Theorem: Suppose $U \opensubset \R^n$ and $f: U \to \R^n$ is smooth. Let $\vva,\vvb \in U$ such that $[\vva,\vvb] \subseteq U$. (Here, $[\vva,\vvb]$ is the line segment joining $\vva$ to $\vvb$.) If there exists $M \in \R$ such that $\norm{Df(\vx)} \le M$, then $\norm{f(\vvb) - f(\vva)} \le M \norm{\vvb - \vva}$.

Lecture 05 (2016-01-25)

Inverse Function Theorem: Let $\vp \in U \opensubset \R^n$ and suppose $f: U \to \R^n$ has the property that $Jf(\vp)$ is invertible. Then $f$ is smoothly invertible near $f(\vp)$. More precisely, there exists $V \opensubset \R^n$ such that $\vp \in V \subseteq U$, $f(V) \opensubset \R^n$, and $f|_V: V \bijto f(V)$ is a bijection. Moreover, its inverse $g: f(V) \to V$ is smooth.

Proof sketch: By translation, we can take $\vp = f(\vp) = \vo$ without loss of generality. By replacing $f$ by $Df(\vp)^{-1} \circ f$, we can take $[Df(\vp)] = I_n$. We will show that there exists $\eps > 0$ such that $B(\vo, \eps) \subseteq U$ and $f$ is injective on $B(\vo, \eps)$. We will then take $V = B(\vo, \eps)$ and do a little more work to show that $f(V)$ is open and $g$ is smooth.

Since $f$ is smooth, whenever $\vx$ is close to $\vo$, the entries of $[Df(\vx)]$ will be close to those of $I_n$. It follows that there exists $r>0$ for which $\norm{\vx} \le r$ implies $\vx \in U$ and $\norm{Df(\vx) - Df(\vo)} < \frac{1}{2}$.

We claim that for all $\vq \in B(\vo, \frac{r}{2})$, there exists a unique $\vvb \in B(\vo, r)$ such that $f(\vvb) = \vq$. To see this, it suffices to show that $F: \overline{B(\vo, r)} \to \R^n$ defined by $F(\vx) = \vx + \vq - f(\vx)$ a unique fixed point $\vvb$. For this, we wish to apply Banach's CMP, so we must show that $F$ is a contraction mapping taking $\overline{B(\vo, r)}$ to $\overline{B(\vo, r)}$.

Observe that

\[\norm{[DF(\vx)]} = \norm{I_n - [Df(\vx)]} < \frac{1}{2} \]

for all $\vx \in \overline{B(\vo, r)}$. Moreover,

\[\norm{F(\vx)} = \norm{\vx + \vq - f(\vx)} \le \norm{\vx - f(\vx)} + \norm{\vq} \le \norm{\vx - f(\vx)} + \frac{r}{2}. \]

Now, by MMVT,

\[\norm{\vx - f(\vx)} = \norm{F(\vx) - F(\vo)} \le \frac{1}{2} \norm{\vx - \vo} \le \frac{r}{2}. \]


\[\norm{F(\vx)} \le \norm{\vx - f(\vx)} + \frac{r}{2} \le \frac{r}{2} + \frac{r}{2} = r. \]

This means that for all $\vx,\vy \in \overline{B(\vo, r)}$, we have

\[\norm{F(\vx) - F(\vy)} \le \frac{1}{2} \norm{\vx - \vy} \]

so $F$ is a contraction mapping. (Extra work to show that $f(V)$ is open and $g$ is smooth omitted.) EOS

Lecture 06 (2016-01-27)

One-Dimensional Integration. Given a continuous function $f: [a,b] \to \R$, we define

\[\int_a^b f(x) \dd{x} = \lim_{k \to \infty} \sum_{i=1}^k f(x_i^*) \Delta_x \]

where $a = x_0 < x_1 < x_2 < \dots < x_k = b$ is a partition of $[a,b]$ into $k$ subintervals of equal length $\Delta x = (b-a)/k$, and $x_i^* \in [x_{i-1}, x_i]$ are arbitrary sample points from these intervals.


Double Integrals. Given a rectangle $R \subseteq \R^2$ and a bounded function $f: R \to \R$, we define $\int_R f$ (sometimes written $\int_R f \dd{A}$) as follows: for each product partition $\PP = \PP_1 \times \PP_2$ of $R = [a,b] \times [c,d]$, let

\[M_{ij} = \sup_{\vx \in R_{ij}} f(\vx) \qquad m_{ij} = \inf_{\vx \in R_{ij}} f(\vx) \]

where $R_{ij}$ is the $(i,j)$th subrectangle of $R$ in $\PP$. Then we define the upper and lower sums of $f$ according to $\PP$ as

\[U(f,\PP) = \sum_{i,j} M_{ij} \operatorname{area}(R_{ij}) \qquad L(f,\PP) = \sum_{i,j} m_{ij} \operatorname{area}(R_{ij}). \]

Definition: We say that $f$ is integrable on $R$ iff there exists a unique number $I$ such that for every partition $\PP$ of $R$,

\[L(f,\PP) \le I \le U(f,\PP). \]

If so, we define $\int_R f \coloneqq I$ and call it the Riemann integral of $f$ on $R$.

Example: Let $R = [0,1] \times [0,1] \subseteq \R^2$, and let $f: R \to \R$ be defined by

\[f(x,y) = \begin{cases} 0 & \text{ if } 0 \le y \le \frac{1}{2} \\ 3 & \text{ if } \frac{1}{2} < y \le 1 \\ \end{cases} \]

We would guess $\int_R f = 3/2$. Consider the partition $\PP = \{R_{11}, R_{12}\}$ of $R$ into two rectangles across $y = \frac{1}{2}$. Then

\[U(f,\PP) = 0\operatorname{area}(R_{11}) + 3\operatorname{area}(R_{12}) = \frac{3}{2} \]


\[L(f,\PP) = 0\operatorname{area}(R_{11}) + 0\operatorname{area}(R_{12}) = 0. \]

Lecture 07 (2016-01 29)

A First Glimpse of Fubini and the Gap Dichotomy. Fubini's theorem allows us to reduce multiple integrals to iterated integrals.

Fubini's Theorem: If $R = [a,b] \times [c,d] \subseteq \R^2$ and $f: \R \to \R$ is continuous, then $f$ is integrable on $R$, and

\[\int_R f = \int_c^d \int_a^b f(x,y) \dd{x} \dd{y}. \]

Example: Let $R = [1,3] \times [2,4]$ and $f(x,y) = x/y$. Then

\[\int_R f = \int_2^4 \int_1^3 \frac{x}{y} \dd{x} \dd{y} = \int_2^4 \frac{4}{y} \dd{y} = 4 \log 2. \]

Definition: Let $R = [a_1,b_1] \times \dots \times [a_n,b_n] \subseteq \R^n$. A partition $\PP$ of $R$ is a tuple $\PP = (\PP_1, \PP_2, \dots, \PP_n)$ of one-dimensional partitions, where $\PP_i$ is a partition of $[a_i,b_i]$. We define the volume of $R$ to be $\vol(R) = \prod_i (b_i - a_i)$.

{$\bm{n}$-dimensional integration.} Similarly to the 2D case, we let $\PP$ be a partition of a rectangle $R \subseteq \R^n$. Then for each $S \in \PP$, we define

\[m_S = \inf_{\vx \in S} f(\vx) \qquad M_S = \sup_{\vx \in S} f(\vx). \]

We then define the lower and upper sums

\[L(f,\PP) = \sum_{S \in \PP} m_S \vol(S) \qquad U(f,\PP) = \sum_{S \in \PP} M_S \vol(S). \]

Definition: $f$ is integrable on $R$ iff there exists a unique $I \in \R$ such that for every partition $\PP$ of $R$,

\[L(f,\PP) \le I \le U(f,\PP). \]

When this occurs, we define $\int_R f \coloneqq I$.

(Example of a non-integrable function omitted. Take the characteristic function of $\Q$.)

The Gap Dichotomy. For any function $f: R \to \R$, let $\mathcal{L}$ be the collection of lower sums of $f$ over $R$, and let $\mathcal{U}$ be the collection of upper sums. Then either there is a gap between $\mathcal{L}$ and $\mathcal{U}$, in which case $f$ is not integrable, or $\mathcal{L}$ and $\mathcal{U}$ touch, and $f$ is integrable.

Corollary: If there exists a sequence $\{\PP_n\}_{n=1}^\infty$ of partitions of $R$ such that

\[\lim_{n \to \infty} L(f, \PP_n) = \lim_{n \to \infty} U(f, \PP_n), \]

then $f$ is integrable over $R$, and $\int_R f$ is that common limit.

Definition: Let $\PP$ and $\PP'$ be partitions of $R$. We say that $\PP'$ is a refinement of $\PP$ if every subrectangle $Q \in \PP'$ is contained in some subrectangle $S \in \PP$. This means that $\PP'$ is obtained from $\PP$ by adding extra endpoints to $\PP$.

Proof of Gap Dichotomy: Observe that if $\PP'$ is a refinement of $\PP$, then

\[L(f,\PP) \le L(f,\PP') \le U(f,\PP') \le U(f,\PP) \]

Furthermore, if $\PP_1$ and $\PP_2$ are two partitions, then they have a common refinement $\PP'$. (Simply draw $\PP_1$ and $\PP_2$ on top of each other.) Then

\[L(f,\PP_1) \le L(f,\PP') \le U(f,\PP') \le U(f,\PP_2). \]

This means that for any two partitions $\PP_1$ and $\PP_2$, we have $L(f,\PP_1) \le U(f,\PP_2)$, as desired. QED

Lecture 08 (2016-02-01)

Integration over Bounded Sets and Volume. Let $\Omega \subseteq \R^n$ be bounded, and let $f: \Omega \to \R$ be a bounded function. Then there exists a rectangle $R \supseteq \Omega$, and we can extend $f$ to $R$ by defining

\[\tilde{f}(\vx) = \begin{cases} f(\vx) & \vx \in \Omega \\ 0 & \vx \in R \setminus \Omega \end{cases} \]

This allows us to define integrals over non-rectangular bounded regions. Provided the RHS exists, we define

\[\int_\Omega f \coloneqq \int_R \tilde{f}. \]

Warning: The RHS might fail to exist even for really “nice” functions $f$. For example, take $\Omega = \Q \cap [0,1]$, and let $f: \Omega \to \R$ be the constant function 1. Then

\[\int_\Omega f = \int_{[0,1]} f \]

fails to exist. This tells us that “niceness” of $f$ is not sufficient to guarantee integrability; we also need “niceness” of the region $\Omega$.

A Second Glimpse of Fubini. Suppose we have two functions $g_1,g_2 : [a,b] \to [c,d]$ for which $g_1(x) \le g_2(x)$ for all $x \in [a,b]$. Let $R = [a,b] \times [c,d]$, and let $\Omega \subseteq R$ be the region bounded by the graphs of $g_1$ and $g_2$. Then by Fubini's theorem,

\[\int_\Omega f \coloneqq \int_R \tilde{f} = \int_a^b \int_c^d \tilde{f}(x,y) \dd{y} \dd{x} = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \dd{y} \dd{x} \]

Example: Find $\int_\Omega (6x + 2y^2)$ where $\Omega$ is the region in $\R^2$ bounded by $x=y^2$ and $x+y=2$. (Solution omitted.)

Definition: Let $\Omega \subseteq \R^n$ be bounded. The $n$-dimensional volume of $\Omega$ is

\[\vol_n(\Omega) \coloneqq \int_\Omega 1 \]

if the RHS exists. (Some discussion omitted.)

Lecture 09 (2016-02-03)

(Discussion of the definition of $n$-dimensional volume omitted.)

Theorem: $\vol_n(X) = 0$ iff for every $\eps > 0$, there exists a partition $\PP$ of $R \supset X$ such that

\[\sum_{\mathclap{S \in \PP, S \cap X \ne \varnothing}} \vol_n(S) < \eps. \]

Proof ($\implies$): Let $\chi$ be the characteristic function of $X$. Given $\eps > 0$, the gap dichotomy implies the existence of a partition $\PP$ such that

\[\sum_{\mathclap{S \in \PP, S \cap X \ne \varnothing}} \vol_n(S) = U(\chi,\PP) < \eps \]

as desired.

($\impliedby$): We first show that $\chi$ is integrable. Let $\eps > 0$ be given. By hypothesis, there exists a partition $\PP$ for which

\[\sum_{\mathclap{S \in \PP, S \cap X \ne \varnothing}} \vol_n(S) < \eps. \]

It follows that

\[U(\chi, \PP) - L(\chi, \PP) = \sum_{\mathclap{S \in \PP, S \cap X \ne \varnothing}} \vol_n(S) - \sum_{\mathclap{S \in \PP, S \subseteq X}} \vol_n(S) \le \sum_{\mathclap{S \in \PP, S \cap X \ne \varnothing}} \vol_n(S) < \eps. \]

It follows that the upper and lower sums of $\chi$ get arbitrarily close; hence, $\int_R \chi$ exists. Moreover, $\inf_\PP U(\chi, \PP) = 0$, so $\int_R \chi = 0$, as desired. QED

Corollary: If $\vol_n(X) = 0$, then $X$ does not contain any $n$-dimensional rectangle.

Theorem: Let $R \subseteq \R^n$ be a rectangle and let $f: R \to \R$ be continuous. Then $\int_R f$ exists.

Proof: Let $\eps > 0$ be given. By the gap dichotomy, it suffices to find a partition $\PP$ of $R$ such that

\[U(f,\PP) - L(f,\PP) < \eps. \]

Because $R$ is compact, $f$ is uniformly continuous on $R$, and there exists $\delta > 0$ such that

\[\norm{\vx - \vy} < \delta \implies \abs{f(\vx) - f(\vy)} < \frac{\eps}{\vol_n(R)} \]

for all $\vx,\vy \in R$. Choose $\PP$ such that every subrectangle $S \in \PP$ has side lengths strictly less than $\delta/\sqrt{n}$. This guarantees that the diameter of $S$ is less than $\delta$, so that

\[\sum_{S \in \PP} (M_s - m_s) \vol_n(S) < \frac{\eps}{\vol_n(R)} \sum_{S \in \PP} \vol_n(S) = \eps \]

as desired. QED

Lecture 10 (2016-02-05)

Fubini's Theorem. Let $f$ be integrable in $R = A \times B \subseteq \R^2$. We want to show that

\[\int_R f = \int_A \int_B f(x,y) \dd{y} \dd{x}. \]

Notation: For all $x \in A$, define $g_x: B \to \R$ by

\[g_x(y) = f(x,y) \]

for all $y$ in $B$. Assume further that for all $x \in A$, $g_x$ is integrable on $B$, and define $G: A \to \R$ by

\[G(x) = \int_B g_x(y) \dd{y} = \int_B g_x. \]

This is the inner integral in the statement of Fubini's theorem.

Theorem: $G$ is integrable on $A$, and

\[\int_R f = \int_A g \]

Proof: Let $\PP$ be a partition of $R$. We can write $\PP = (\PP_A, \PP_B)$ where $\PP_A$ and $\PP_B$ are partitions of $A$ and $B$, respectively. We will show that

\[L(f,\PP) \le L(G,\PP_A) \le U(G,\PP_A) \le U(f,\PP). \]

This will prove the theorem by the gap dichotomy. Indeed, we are given that no gap exists between $L(f,\PP)$ and $U(f,\PP)$. The preceding inequality will show that no gap exists between $L(G,\PP_A)$ and $U(G,\PP_A)$. Let

\[\begin{aligned} \PP_A &= \{T_1, T_2, \dots, T_{k_1}\} \\ \PP_B &= \{U_1, U_2, \dots, U_{k_2}\}. \end{aligned} \]

Then every $S \in \PP$ is of the form $T_i \times U_j$ for some $(i,j)$, and we can write

\[L(f,\PP) = \sum_{S \in \PP} m_S(f) \vol_2(S) = \sum_{i=1}^{k_1} \sum_{j=1}^{k_2} m_{T_i \times U_j}(f) \vol_1(T_i) \vol_1(U_j). \]

Now note that for all $x \in T_i$, we have

\[m_{T_i \times U_j}(f) \le m_{U_j}(g_x). \]

It follows that for all $1 \le i \le k_1$ and $x \in T_i$, we have

\[\sum_{j=1}^{k_2} m_{T_i \times U_j}(f) \vol_1(U_j) \le \sum_{j=1}^{k_2} m_{T_i \times U_j}(g_x) \vol_1(U_j) = L(g_x,\PP) \le \int_B g_x = G(x) \]

Now multiply both sides by $\vol_1(T_i)$ and sum over $i$ to obtain

\[\sum_{i=1}^{k_1} \sum_{j=1}^{k_2} m_{T_i \times U_j}(f) \vol_1(U_j) \vol_1(T_i) \le \sum_{i=1}^{k_1} m_{T_i}(G) \vol_1(T_i) = L(G,\PP_A). \]

This shows that

\[L(f,\PP) \le L(G,\PP_A), \]

as desired. The inequality for the upper sums follows similarly. QED

Lecture 11 (2016-02-08)

Introduction to Change of Variables. We will often find that a multiple integral can be broken into iterated integrals more naturally in a non-Cartesian coordinate system. For example, the area of a circular sector is easier to calculate in polar coordinates.

Polar Coordinates $(r,\theta)$. These coordinates are defined by

\[x = r \cos\theta \qquad y = r \sin\theta. \]

Let $g: \R^2 \times \R^2$ be the coordinate transformation function

\[g(r,\theta) = \mqty[r\cos\theta \\ r\sin\theta]. \]

We usually restrict $g: [0,\infty) \times [0,2\pi) \to \R^2$. Note that

\[\det Jg(r,\theta) = \det\mqty[\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta] = r. \]

Area Distortion. Let $X$ be a domain in the $(r,\theta)$-plane and let $\Omega = g(X)$. In general, we will find

\[\int_\Omega f \ne \int_X f \circ g \]

because the coordinate transformation $g$ does not preserve area. We need to multiply the RHS by $\det Jg$ to compensate for this.

\[\int_\Omega f \ne \int_X (f \circ g) \det Jg \]

(Example omitted.)

Cylindrical Coordinates in $\R^3$. Let $g: [0,\infty) \times [0,2\pi) \times \R \to \R^3$ be defined by

\[g(r,\theta,z) = \mqty[r\cos\theta \\ r\sin\theta \\ z]. \]

Then $\det Jg(r,\theta,z) = r$.

(Example omitted.)

Lecture 12 (2016-02-10)

Discussion of test review omitted.

Lecture 13 (2016-02-12)

Spherical Coordinates. This is the coordinate system $(\rho, \phi, \theta)$ defined by $g: [0,\infty) \times [0,\pi] \times [0,2\pi) \to \R^3$, where

\[\begin{aligned} x &= \rho\sin\phi\cos\theta \\ y &= \rho\sin\phi\sin\theta \\ z &= \rho\cos\phi. \end{aligned} \]

(Note that $\theta$ is the azimuthal angle, and $\phi$ is the polar angle.) The determinant of this transformation is

\[\det[Jg(\rho,\phi,\theta)] = \rho^2 \sin\phi \]

(Example omitted: volume of sphere.)

Lecture 14 (2016-02-15)

Definition: An $n \times n$ determinant is an alternating multilinear function $\mathcal{D}: \underbrace{\R^n \times \R^n \times \cdots \times \R^n}_{n \text{ times}} \to \R$ with the normalization property $\mathcal{D}(\ve_1, \dots, \ve_n) = 1$.

Theorem: For every $n \in \N$, there exists a unique $n \times n$ determinant.

Proof: The theorem is clear in the $1 \times 1$ case: for a scalar $a$, we must have $\mathcal{D}(a) = a$.

To demonstrate existence for higher $n$, we can inductively define $\det: \mathbb{M}_n \to \R$ as follows: for $A \in \mathbb{M}_n$, let $M_{ij} \in \mathbb{M}_{n-1}$ be the matrix obtained by deleting the $i$th row and $j$th column of $A$. Let $c_{ij} \coloneqq (-1)^{i+j} \det M_{ij}$ be the cofactors of $A$. Then we define

\[\det A = \sum_{i=1}^n A_{i1} c_{i1} = A_{11}c_{11} + A_{21}c_{21} + \dots + A_{n1}c_{n1}. \]

The function $\det$ defined in this way (so-called “cofactor expansion”) is an $n \times n$ determinant. (We omit details of the verification here.)

The proof of uniqueness requires the following 12-step program: Let $\mathcal{D}: \mathbb{M}_n \to \R$ be an $n \times n$ determinant.

  1. If $A \in \mathbb{M}_n$ has two distinct equal columns, then $\mathcal{D}(A) = 0$.

  2. If $B \in \mathbb{M}_n$ is obtained from $A$ by multiplying one of its columns by $c$, then $\mathcal{D}(B) = c\mathcal{D}(A)$.

  3. If $\vo$ is a column of $A$, then $\mathcal{D}(A) = 0$.

  4. If $B$ is obtained from $A$ by adding a scalar multiple of column $i$ to column $j \ne i$, then $\mathcal{D}(B) = \mathcal{D}(A)$.

Lecture 15 (2016-02-17)


Lecture 15.5 (2016-02-18)

Theorem: If $A \in \mathbb{M}_n$, then $\det(A) = \det(A^\intercal)$.

Proof: If $\rank(A) < n$, then $\det(A) = \det(A^\intercal) = 0$ since $\rank(A) = \rank(A^\intercal)$.

If $\rank(A) = n$, write $A = E_1E_2\cdots E_m$ as a product of elementary matrices. It is easy to show by cases that $\det(E) = \det(E^\intercal)$. Hence, $\det(A) = \det(A^\intercal)$ by the product rule. QED

Theorem: If $X \in \R^n$ has an $n$-dimensional volume and $T: \R^n \to \R^n$ is a linear transformation, then $T(X)$ has $n$-dimensional volume, and

\[\vol_n(T(X)) = \abs{\det[T]} \vol_n(X). \]

Proposition: Let $R = [a_1,b_1] \times [a_2,b_2] \times \dots \times [a_n,b_n]$ be a rectangle in $\R^n$ and $T: \R^n \to \R^n$ be a linear transformation. Then

\[\vol_n(T(R)) = \begin{cases} 0 & \text{if $[T]$ is not invertible} \\ \vol_n(R) & \text{if $[T]$ is elementary of type I} \\ \abs{c} \vol_n(R) & \text{if $[T]$ is elementary of type II$_c$} \\ \vol_n(R) & \text{if $[T]$ is elementary of type III} \end{cases} \]

Proof: If $[T]$ is not invertible, then $\operatorname{Image}(T)$ is a linear subspace of $\R^n$ of dimension $k < n$. Any bounded subset of $\operatorname{Image}(T)$ thus has volume $0$.

If $[T]$ is elementary of type I or II$_c$, then $T$ takes rectangles to rectangles, and the theorem is established with simple casework. (We omit details here.)

If $[T]$ is elementary of type III, then $[T]$ has one nonzero off-diagonal entry. Say $[T]$ performs the elementary row operation $R_j \mapsto R_j + cR_i$, so that the nonzero entry $c$ occurs at location $(j,i)$. Then the region $T(R)$ is bounded by

\[\begin{aligned} a_1 \le &y_1 \le b_1 \\ &\vdots \\ a_i \le &y_i \le b_i \\ cy_i + a_j \le &y_j \le cy_i + b_j \\ &\vdots \\ a_n \le &y_n \le b_n \end{aligned} \]

and a simple application of Fubini's theorem finishes the proof. QED

Lecture 16 (2016-02-19)

Proposition: Let $T: \R^n \to \R^n$ be a linear transformation, and suppose $X \subseteq \R^n$ has $n$-dimensional volume. Then $\vol_n(T(X)) = \abs{\det[T]}\vol_n(X)$, provided $T$ is either singular or elementary.

Proof: Let $X \subseteq R \subseteq \R^n$ for some rectangle $R$, so that $\vol_n(X) = \int_X 1 = \int_R \chi$. For any partition $\PP$ of $R$, we have

\[\vol_n(X) \le \mathcal{U}(\chi, \PP) = \sum_{\mathclap{\substack{S \in \PP \\ S \cap X \ne \varnothing}}} \vol_n(S). \]

It follows, by the preceding proposition, that

\[\begin{aligned} \vol_n(T(X)) &\le \sum_{\mathclap{\substack{S \in \PP \\ S \cap X \ne \varnothing}}} \vol_n(T(S)) \\ &= \sum_{\mathclap{\substack{S \in \PP \\ S \cap X \ne \varnothing}}} \abs{\det[T]} \vol_n(S) \\ &= \abs{\det[T]} \sum_{\mathclap{\substack{S \in \PP \\ S \cap X \ne \varnothing}}} \vol_n(S) \\ &= \abs{\det[T]} \mathcal{U}(\chi, \PP) \end{aligned} \]

A similar argument for lower sums shows that

\[\abs{\det[T]} \mathcal{L}(\chi, \PP) \le \vol_n(T(X)) \le \abs{\det[T]} \mathcal{U}(\chi, \PP). \]

Now, either $\det[T] = 0$, in which case $\vol_n(T(X)) = 0$, $\det[T] \ne 0$, in which case

\[\mathcal{L}(\chi, \PP) \le \frac{\vol_n(T(X))}{\abs{\det[T]}} \le \mathcal{U}(\chi, \PP) \implies \vol_n X = \frac{\vol_n(T(X))}{\abs{\det[T]}}. \]


Theorem: Let $T: \R^n \to \R^n$ be a linear transformation, and suppose $X \subseteq \R^n$ has $n$-dimensional volume. Then $\vol_n(T(X)) = \abs{\det[T]}\vol_n(X)$.

Proof: If $[T]$ is singular, then both sides are zero by the preceding proposition. If $[T]$ is invertible, write $T$ as a product of elementary matrices and repeatedly apply the preceding proposition using the product rule for determinants. QED

The Linear Change of Variables Theorem: Let $T: \R^n \to \R^n$ be a linear transformation, $X \subseteq \R^n$, and $f: T(X) \to \R$ be bounded. Then

\[\int_{T(X)} f = \int_X f \circ T \abs{\det[T]} \]

assuming either integral exists.

Proof sketch (assuming both integrals exist): For simplicity of notation, let

\[g(\vx) = f(T(\vx)) \abs{\det[T]} \]

and assume $X$ is a rectangle. Let $\PP$ be a partition of $X$. Then

\[\begin{aligned} \int_X g &\le \mathcal{U}(g, \PP) \\ &= \sum_{S \in \PP} M_S(g) \vol_n(S) \\ &= \sum_{S \in \PP} M_S(f \circ T) \abs{\det[T]} \vol_n(S) \\ &= \sum_{S \in \PP} M_S(f \circ T) \vol_n(T(S)) \\ &= \sum_{S \in \PP} M_{T(S)}(f) \vol_n(T(S)) \\ &\ge \sum_{S \in \PP} \int_{T(S)} f = \int_{T(X)} f. \end{aligned} \]

A similar argument for lower sums shows that $\int_{T(X)} f$ is between every upper and lower sum of $\int_X g$, so the two are equal. QED

The General Change of Variables Theorem: For $g$ smooth,

\[\int_{g(X)} f = \int_X (f \circ g) \abs{\det Jg}. \]

Lecture 17 (2016-02-22)

{Parallelepipeds and $\bm{k}$-dimensional volume.} Let $\vva_1, \dots, \vva_k$ be linearly independent vectors in $\R^n$.

Definition: The $k$-dimensional parallelepiped spanned by $\vva_1, \dots, \vva_k$ is

\[P(\vva_1, \dots, \vva_k) = \qty{\sum_{i=1}^k t_i\vva_i : 0 \le t_i \le 1}. \]

For $k = n$, we define the unit $n$-cube $Q^n = P(\ve_1, \dots, \ve_n)$.

Let $A = \mqty[\vva_1 & \cdots & \vva_k] \in \mathbb{M}_{n,k}$. Then $P(\vva_1, \dots, \vva_k) = A(Q^k)$. If $k = n$, we define

\[\vol_n P(\vva_1, \dots, \vva_n) = \abs{\det A} \]

which is motivated by our Linear Change of Variables Theorem.

{Volume with $\bm{k < n}$.} If $V$ is a linear subspace of $\R^n$ and $X \subseteq V$ is bounded, we would like to have a reasonable definition of $\vol_k(X)$. We will take the following approach:

  1. Find a linear isometry $V \to \R^k$, that is, a linear transformation $T: \R^n \to \R^k$ such that $T|_V$ is an isometry.

  2. Define $\vol_k(X) \coloneqq \vol_k(T(X))$.

  3. Prove that this definition is independent of the choice of $T$.

  4. Theorem: If $A = \mqty[\vva_1 & \cdots & \vva_k] \in \mathbb{M}_{n,k}$ has rank $k$, then

    \[\vol_k P(\vva_1, \dots, \vva_k) = \sqrt{\det(A^\intercal A)}. \]

    Note that for $n = k$, this reduces to $\vol_k P(\vva_1, \dots, \vva_n) = \abs{\det A}$, as before.

Recall: $A \in \mathbb{M}_n$ is orthogonal if $A^\intercal A = I_n$, that is, if the columns of $A$ form an orthonormal basis.

Theorem: If $V \subseteq \R^n$ is a $k$-dimensional linear subspace, then $V$ has an orthonormal basis.

Proof: Build up the basis incrementally. Take any unit vector $\vvb_1$, take $\vvb_2 \in V \cap \operatorname{span}\{\vvb_1\}^\perp$, and so on. QED

Thus, we can always find $B = \mqty[\vvb_1 & \cdots & \vvb_k] \in \mathbb{M}_{n,k}$ with $C(B) = V$, $\rank(B) = k$, and $B^\intercal B = I_k$.

Lecture 18 (2016-02-24)

Proposition: If $B \in \mathbb{M}_{n,k}$ satisfies $B^\intercal B = I_k$ (i.e., the columns of $B$ are orthonormal), then $B^\intercal$ preserves dot products of vectors in $V = C(B)$. That is, for all $\vx, \vy \in V$, we have $\vx \cdot \vy = B^\intercal \vx \cdot B^\intercal \vy$.

Proof: Since $\vx, \vy \in C(B)$, we can write $\vx = B\vz$ and $\vy = B\vw$ for some $\vz, \vw \in \R^k$. It follows that

\[\vx \cdot \vy = B\vz \cdot B\vw = B^\intercal B \vz \cdot \vw = \vz \cdot \vw = B^\intercal \vx \cdot B^\intercal \vy \]

as desired. QED

Corollary: If $B \in \mathbb{M}_{n,k}$ satisfies $B^\intercal B = I_k$, then $B^\intercal$ is an isometric embedding of $V$ in $\R^k$.

Definition: If $X$ is a subset of a $k$-dimensional linear subspace $V$ of $\R^n$, then we define $\vol_k(X) = \vol_k(B^\intercal(X))$, where $B^\intercal$ is an isometric embedding of $V$ in $\R^k$. (By the preceding proposition, such an embedding always exists by taking the columns of $B$ to be an orthonormal basis of $V$.)

The following theorem shows that $\vol_k$ is well-defined.

Theorem: Let $A \in \mathbb{M}_{n,k}$ have rank $k$. Let $Y \subseteq \R^k$ be bounded. Then

\[\vol_k(A(Y)) = \sqrt{\det(A^\intercal A)} \vol_k(Y). \]

Proof: Let $B \in \mathbb{M}_{n,k}$ be a matrix whose columns form an orthonormal basis for $C(A)$. By definition,

\[\vol_k(A(Y)) = \vol_k(B^\intercal A(Y)) = \abs{\det(B^\intercal A)} \vol_k(Y) \]

where we note that $B^\intercal A$ is square. Thus, it suffices to show that

\[\det(B^\intercal A)^2 = \det(A^\intercal A). \]

Observe that

\[\begin{aligned} \det(B^\intercal A)^2 &= \det(B^\intercal A) \det(B^\intercal A) \\ &= \det(A^\intercal B) \det(B^\intercal A) \\ &= \det(A^\intercal B) \det(B^\intercal A) \\ &= \det(A^\intercal B B^\intercal A). \end{aligned} \]

We claim that $A^\intercal B B^\intercal A = A^\intercal A$. To see this, let $\vx \in \R^k$ be arbitrary. Then there exists $\vy \in \R^k$ such that $A\vx = B\vy$. It follows that

\[A^\intercal B B^\intercal A \vx = A^\intercal B B^\intercal B \vy = A^\intercal B \vy = A^\intercal A \vx. \]

Since this holds for arbitrary $\vx \in \R^k$, we have $A^\intercal B B^\intercal A = A^\intercal A$, as desired. This completes the proof. QED

Corollary: If $A \in \mathbb{M}_{n,k}$ has rank $k$, then $\det(A^\intercal A) > 0$.

Proof: We have already seen that $\det(A^\intercal A)$ is nonnegative, so it suffices to show that $\det(A^\intercal A)$ is nonzero. Suppose $A^\intercal A \vx = \vo$. Then

\[0 = A^\intercal A \vx \cdot \vx = A\vx \cdot A\vx \implies A\vx = \vo \implies \vx = \vo \]

where the last implication follows from the rank-nullity theorem. Thus, $A^\intercal A$ is invertible, and $\det(A^\intercal A) \ne 0$. QED

Corollary: If $A = \mqty[\vva_1 & \cdots & \vva_k]$ has rank $k$, then

\[\vol_k P(\vva_1, \dots, \vva_k) = \sqrt{\det(A^\intercal A)}. \]

Lecture 19 (2016-02-26)

Integration of Scalar-Valued Functions on Manifolds. Let $X$ be a bounded subset of an open set $U$ in $\R^k$, and let $g: U \to \R^n$ be smooth. Then $g(X)$ is a $k$-parameterized region in $\R^n$. We would like to construct a reasonable definition for the integral of $f$ over $g(X)$.

Design Criteria:

  1. If $f = 1$ is a constant function and $g$ is injective, our definition should give the $k$-dimensional volume of $g(X)$.

  2. Our definition should be consistent with previous theorems. In particular,

  3. If $g$ is a linear transformation and $f = 1$, we should get $\vol_k(g(X))$.

  4. If $n=k$ and $g = \operatorname{id}_U$, we should get $\int_X f$.

Based on these criteria, we can rule out a few possible candidates:

Definition: The integral of $f$ on $g(X)$ with respect to $k$-dimensional volume is

\[\int_{g(X)} f \dd{V_k} \coloneqq \int_X (f \circ g) \sqrt{\det(Jg^\intercal Jg)}. \]

Remark: We often write $\dd{V}$ for $\dd{V_3}$, $\dd{A}$ for $\dd{V_2}$, and $\dd{s}$ or $\abs{\dd{x}}$ for $\dd{V}_1$.

(Omitted discussion of physical interpretation of this integral as measuring total mass for a given density distribution.)

Lecture 20 (2016-02-29)

_{(Missed proof that $\vol_k$ is well-defined. Show that $\vol_k(B^\intercal X) = \vol_k(A^\intercal X)$ for any two matrices $A, B$ whose columns form an orthonormal basis of $V$. Let $Y,Z \subseteq \R^k$ such that $AY = BZ = X$. Then

\[\vol_k(Z) = \vol_k(B^\intercal X) = \vol_k(X) = \sqrt{\det(A^\intercal A)} \vol_k(Y) = \vol_k(Y) \]

as desired.)}_

Recall our new integral

\[\int_{g(X)} f \dd{V}_k \coloneqq \int_X (f \circ g) \sqrt{\det(Jg^\intercal Jg)}. \]

If $g$ is injective on $X$, then

\[\vol_k(g(X)) \coloneqq \int_{g(X)} 1 \dd{V}_k = \int_X \sqrt{\det(Jg^\intercal Jg)}. \]

Note that in the special case $k=1$, we obtain the usual formula for the arc length of a parametric curve:

\[\vol_1(g([a,b])) = \int_{g([a,b])} \dd{V}_1 = \int_a^b \norm{g'(t)} \dd{t}. \]

(Examples omitted: arc length of conical helix and surface area of unit 2-sphere.)

Lecture 21 (2016-03-02)

(Missed definition of $k$-forms.)

Main Example: Let $I = (i_1, \dots, i_k)$ where $i_1, \dots, i_k$ are integers between $1$ and $n$. we call $I$ a multi-index. Define $\phi: \mathbb{M}_{n,k} \to \R$ as follows: write

\[A = \mqty[A_1^\intercal \\ \vdots \\ A_n^\intercal] \text{, and let } A_I = \mqty[A_{i_1}^\intercal \\ \vdots \\ A_{i_k}^\intercal]. \]

Then we define $\phi(A) = \det(A_I)$.

Notation: We typically give this $k$-form $\phi$ the name $\dd{x}_I$. We call $\dd{x}_I$ an elementary $k$-form.

(Basic examples of computations omitted.)

Fact: If $I = (i_1, \dots, i_k)$, then

\[\dd{x}_I\!\qty(\ve_{i_1}, \dots, \ve_{i_k}) = \begin{cases*} 0 & \text{if two indices are the same} \\ 1 & \text{otherwise}. \end{cases*} \]

Geometric interpretation. Let $A = \mqty[\vv_1 & \cdots & \vv_k]$ and $I = (i_1, \dots, i_k)$. Then the columns of $A$ span a $k$-dimensional parallelepiped $P$ in $\R^n$. Project $P$ to the $k$-dimensional coordinate subspace $\R_I^k \subseteq \R^n$ to obtain a $k$-dimensional parallelepiped $P_I$ in $\R_I^k$. Then $\dd{x}_I\!(A) = \pm \vol_k(P_I)$.

Definition: $\bigwedge\!\!^k(\R^n)$ is the set of all $k$-forms on $\R^n$. (This notation is not universal; other authors may call this $\Lambda^k(\R^{n*})$ or $A^k(\R^n)$, where $A$ stands for “alternating.”) We also define a $0$-form to be a number, so that $\bigwedge\!\!^0(\R^n) = \R$.

Note: $\bigwedge\!\!^k(\R^n) = \{\vo\}$ if $k > n$.

Definition: If $I = (i_1, \dots, i_k)$ is strictly increasing, then $\dd{x}_I$ is called a basic $k$-form.


  1. $\bigwedge\!\!^k(\R^n)$ is a vector space (under the standard addition and scalar multiplication of real-valued functions).

  2. If $0 \le k \le n$, then $\dim \bigwedge\!\!^k(\R^n) = \binom{n}{k}$.

  3. For $1 \le k \le n$, the basic $k$-forms on $\R^n$ form a basis for $\bigwedge\!\!^k(\R^n)$.

Lecture 22 (2016-03-04)

{$\bm{k}$-forms and differential $\bm{k}$-forms.} Recall that $\bigwedge^k(\R^n)$ is the vector space of all $k$-forms on $\R^n$, which are multilinear alternating functions

\[\underbrace{\R^n \times \dots \times \R^n}_{\text{$k$ times}} \to \R. \]

There are $\binom{n}{k}$ basic $k$-forms on $\R^n$. Thus, $\bigwedge^k(\R^n)$ is $\binom{n}{k}$-dimensional.

Proposition: $\displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}$.

(Proof omitted. Elementary combinatorics.)

Lecture 23 (2016-03-14)

Recall that a $k$-form on $\R^n$ is a multilinear alternating function $\varphi: (\R^n)^k \to \R$. We denote by $\bigwedge^k(\R^n)$ the vector space of all $k$-forms on $\R^n$. If $I = (i_1, \dots, i_k)$ is an increasing multi-index (i.e., if $1 \le i_1 < i_2 < \dots < i_k \le n$), then we define the basic $k$-form $\dd{x}_I \in \bigwedge^k(\R^n)$ by $\dd{x}_I(A) = \det(A_I)$, where $A_I$ is the matrix formed by taking rows $i_1, \dots, i_k$ of $A$.

Theorem: If $1 \le k \le n$, then the basic $k$-forms on form a basis for $\bigwedge^k(\R^n)$. Hence, $\dim \bigwedge^k(\R^n) = \binom{n}{k}$.

Special Case: If $k = n$, then $\dim \bigwedge^n(\R^n) = 1$, with $\dd{x}_{(1,2,\dots,n)} = \det$. For each $\varphi \in \bigwedge^n(\R^n)$, there exists $c \in \R$ such that $\varphi = c\det$. In particular, $c = \varphi(I_n)$.

Special Case: If $k = 1$, then the 1-forms $\varphi \in \bigwedge^1(\R^n)$ are precisely the linear transformations $\varphi: \R^n \to \R$. Thus, $\bigwedge^1(\R^n)$ can be viewed as the space of row vectors of length $n$, which clearly has dimension $n$.

Remark: We call $\{ \dd{x}_1, \dots, \dd{x}_n \}$ the dual basis for $\bigwedge^1(\R^n)$.

General case: Given $\varphi \in \bigwedge^k(\R^n)$, we can write $\varphi = \sum_I a_I \dd{x}_I$, where the sum is taken over strictly increasing multi-indices $I$, and

\[a_I = \varphi(\ve_{i_1}, \dots, \ve_{i_k}). \]

It is not difficult to show that this expansion holds for all $k$-forms $\varphi$. We will omit the details here.

Definition: If $\omega \in \bigwedge^k(\R^n)$ and $\eta \in \bigwedge^\ell(\R^n)$, then their wedge product $\omega \wedge \eta \in \bigwedge^{k+\ell}(\R^n)$ is the $(k+\ell)$-form defined by

\[\dd{x}_I \wedge \dd{x}_J = \dd{x}_{(I,J)} \]

for basic forms and extending linearly for other forms. (Example omitted.)

Lecture 24 (2016-03-16)

Review for Exam II. Discussion omitted.

Lecture 25 (2016-03-18)

Missed this lecture.

Lecture 26 (2016-03-21)

The Exterior Derivative, Vector Fields, Work. Let $U \opensubset \R^n$. Recall that $\Omega^k(U)$ denotes the vector space of all differential $k$-forms on $U$, where $\omega \in \Omega^k(U)$ iff for each increasing multi-index $I = (i_1, \dots, i_k)$ there exists a smooth function $f_I: U \to \R$ such that

\[\omega = \sum_I f_I \dd{x}_I. \]

We can evaluate $\omega$ at a point $p$ to obtain an ordinary $k$-form

\[\omega_p = \sum_I f_I(p) \dd{x}_I \in \bigwedge{}^{\!\!\!k}(\R^n). \]

We also define $\Omega^0(U) = C^\infty(U) = $ set of all smooth functions $f: U \to \R$.

The exterior derivative is a map $\dd: \Omega^k(U) \to \Omega^{k+1}(U)$. For $k = 0$, we define $\dd: \Omega^0(U) \to \Omega^1(U)$ by

\[\dd{f} = \pdv{f}{x_1} \dd{x}_1 + \dots + \pdv{f}{x_n} \dd{x}_n. \]

Note that $\dd(x_i) = \dd{x}_i$. The object on the left is the exterior derivative of a smooth function; the object on the right is a basic 1-form.

For $k \ge 1$, we define the exterior derivative by declaring

\[\dd(f_I \dd{x}_I) = \dd{f}_I \wedge \dd{x}_I = \sum_{i=1}^n \pdv{f_I}{x_i} \dd{x}_i \wedge \dd{x}_I \]

and extending linearly.

(Example of differentiating a 1-form omitted.)

Lecture 27 (2016-03-23)

Missed this lecture.

Lecture 28 (2016-03-25)

Let $U$ be an open subset of $\R^n$.

Definition: $F \in VF^\infty(U)$ is conservative iff there exists $f \in C^\infty(U)$ such that $\nabla f = F$. In this case, we say that $f$ is a potential function for $F$. Similarly, $\omega \in \Omega^1(U)$ is exact iff there exists $f \in \Omega^0(U)$ such that $\dd f = \omega$.

Fundamental Theorem of Line Integrals: Let $F$ be a conservative vector field on $U$ with potential function $f$, and let $C$ be a smooth curve in $U$ parameterized by $g: [a,b] \to U$. Then

\[\int_C W_F = f(g(b)) - f(g(a)). \]

Here, $W_F = \dd{f}$ is the work form of $F$, which is obtained from $F$ by replacing unit vectors $\hat{\mathbf{x}}$ by unit 1-forms $\dd{x}_1$.

Proof: Straightforward chasing through definitions. QED

Example: Consider the “toilet bowl” field

\[F(x,y) = \frac{1}{x^2 + y^2} \mqty[-y\\x] \]

defined on $\R^2 \setminus \{\vo\}$. Observe that $F$ is not conservative (since the integral of $F$ over the unit circle is nonzero), but $\dd{W}_F = 0$.

Lecture 29 (2016-03-28)

Missed this lecture.

Lecture 30 (2016-03-30)

Stokes' Theorem: Let $M$ be a compact oriented $(k+1)$-manifold with boundary. Suppose $M \subseteq U \opensubset \R^n$ and $\omega \in \Omega^k(U)$. Then

\[\int_M \dd{\omega} = \int_{\partial M} \omega. \]

Definition: An orientation of $\R^n$ is represented by an ordered basis \linebreak $(\vv_1, \dots, \vv_n)$ of $\R^n$. This orientation is the positive orientation iff

\[\det\mqty[\vv_1 & \cdots & \vv_n] > 0. \]

Otherwise, it is the negative orientation.


Fact: $(\vv_1, \dots, \vv_n)$ represents the positive orientation iff the matrix

\[\mqty[\vv_1 & \cdots & \vv_n] \]

can be column reduced to $I_n$ such that the number $N$ of column exchanges and scalar multiplications of columns by negative scalars is even.

Fancier way: Choose a nonzero $\eta \in \bigwedge^{\!\!n}(\R^n)$. Then the bases $(\vv_1, \dots, \vv_n)$ and $(\vw_1, \dots, \vw_n)$ represent the same orientation iff $\eta(\vv_1, \dots, \vv_n)$ and \linebreak $\eta(\vw_1, \dots, \vw_n)$ have the same sign.

Definition: Let $M$ be a $k$-dimensional manifold in $\R^n$. An orientation form for $M$ is a differential form $\eta \in \Omega^k(U)$ defined on some neighborhood $U$ of $M$ such that $\eta$ is non-vanishing on $M$. This means $\forall \vp \in M$ $\exists \vv_1, \dots, \vv_k \in T_{\vp} M$ such that $\eta_{\vp}(\vv_1, \dots, \vv_k) \ne 0$.

Fact: This is true iff $\eta_{\vp}(\vv_1, \dots, \vv_k) \ne 0$ for every basis $(\vv_1, \dots, \vv_k)$ of $T_{\vp} M$.

Example: Consider $S^2 = \{\vp \in \R^3 : \norm{\vp} = 1\}$. Define $\eta \in \Omega^2(\R^3)$ by

\[\eta_\vp(\vv_1, \vv_2) = \det\mqty[\vp & \vv_1 & \vv_2]. \]

This an orientation form on $S^2$, since $\vp$ is always orthogonal to the plane $T_\vp S^2$. This shows that $S^2$ is orientable.

Alternatively, recall that the flux form $\Phi_F \in \Omega^{n-1}(\R^n)$ of a vector field $F \in VF^\infty(\R^n)$ is defined by

\[\Phi_{F(\vp)}(\vv_1, \dots, \vv_{n-1}) = \det\mqty[F(\vp) & \vv_1 & \cdots & \vv_n] \]

so our orientation form is actually the flux form of the identity vector field on $\R^3$.

Lecture 30.5 (2016-03-31)

Orientation of hypersurfaces. Let $M$ be a $k$-dimensional manifold embedded in $\R^n$, where $k = n - 1$.

Definition: A vector field $V$ defined on a neighborhood of $M$ is a normal field on $M$ iff for all $\vp \in M$, $V(\vp)$ is orthogonal to $M$. It is nowhere zero on $M$ iff $V(\vp) \ne \vo$ for all $\vp \in M$.

Theorem: $M$ is orientable iff $M$ admits a nowhere zero normal field.

(Some example computations omitted.)

Lecture 31 (2016-04-01)

(Example computations with Stokes's theorem. Omitted.)

Lecture 32 (2016-04-04)

Stokes's Theorem: If $M$ is a compact oriented $(k+1)$-manifold in $\R^n$ and $\omega \in \Omega^k(M)$, then

\[\int_M \dd{\omega} = \int_{\partial M} \omega \]

where $\partial M$ has the induced orientation compatible with $M$.

What does compatible mean? Suppose $\eta \in \Omega^{k+1}(M)$ and $\beta \in \Omega^k(\partial M)$ are orientations for $M$ and $\partial M$.

Definition: $\beta$ and $\eta$ are compatible iff $\forall \vp \in \partial M$, $\forall$ bases $\vv_1, \dots, \vv_k$ of $T_\vp \partial M$, $\beta_\vp(\vv_1, \dots, \vv_k)$ has the same sign as $\eta_\vp(\mu_\vp, \vv_1, \dots, \vv_k)$, where $\mu_\vp = Dg(\vq)(-\ve_{k+1}) \in T_\vp M$ is an outward-pointing tangent vector to $M$ at $\vp$. Here, $g$ is a boundary chart mapping $\R^k \times [0,\infty)$ to a coordinate neighborhood of $\vp$ in $M$ such that $g(\vq) = \vp$.

Example: $\R^{k+1}_+$ is a $(k+1)$-manifold with boundary in $\R^{k+1}$.

Theorem: If $\eta$ is the standard orientation of $\R_+^{k+1} \subseteq \R^{k+1}$ and $\omega$ is the standard orientation of $\R^k = \partial \R^{k+1}_+$, then $(-1)^{k+1}\omega = \beta$ and $\eta$ are compatible.

Proof: Immediate. QED

Lecture 33 (2016-04-06)

(Some discussion of partitions of unity omitted. To integrate a form over a manifold, pick a coordinate cover and a partition of unity subordinate to it, and integrate the form as usual in coordinates. Sum up the results.)

Proof of Stokes's Theorem: It suffices to prove Stokes's theorem for compact rectangles $R$ in $\R^{k+1}_+$. Indeed, by taking a partition of unity on a manifold with boundary, we reduce integration of a form to integration over coordinate charts and boundary charts.

Given $\omega \in \Omega^k(\R^{k+1}_+)$ where $\omega = 0$ outside a compact rectangle $R$, write

\[\omega = \sum_{i=1}^{k+1} f_i \dd{x}_1 \wedge \dots \wedge \widehat{\dd{x}_i} \wedge \dots \wedge \dd{x}_{k+1} \]

where the hat denotes omission as usual, and $f$ is zero on the boundaries of $R$, except for the boundary of the upper half-space $\R^{k+1}_+$. (This is the boundary that matters in Stokes's theorem.) Then

\[\dd{\omega} = \sum_{i=1}^{k+1} (-1)^{i-1} \pdv{f_i}{x_i} \dd{x}_1 \wedge \dots \wedge \dd{x}_{k+1} \]

and it follows that

\[\begin{aligned} \int_R \dd{\omega} &= \sum_{i=1}^{k+1} (-1)^{i-1} \int_R \pdv{f_i}{x_i} \\ &= \sum_{i=1}^{k+1} (-1)^{i-1} \int_0^{b_{k+1}} \!\! \int_{a_k}^{b_k} \dots \int_{a_1}^{b_1} \pdv{f_i}{x_i} (x_1, \dots, x_{k+1}) \\ &= (-1)^{k+1} \int_{a_k}^{b_k} \dots \int_{a_1}^{b_1} f_{k+1}(x_1, \dots, x_k, 0). \end{aligned} \]

On the other hand,

\[\begin{aligned} \int_{\partial R} \omega &= \int_{\partial R \cap \partial \R^{k+1}_+} \omega \\ &= \sum_{i=1}^{k+1} \int f_i(x_1, \dots, x_k, 0) {\determinant}_i^*\mqty[I_k\\0] \\ &= \int_{[a_1, b_1] \times \dots \times [a_k, b_k]} f_{k+1}(x_1, \dots, x_k, 0) \end{aligned} \]

where ${\determinant}_i^*(A)$ denotes the determinant of $A$ omitting the $i$th row. QED

Lecture 34 (2016-04-08)

Change of Basis. The goal of our next unit will be to represent a linear transformation with respect to a (possibly more convenient) basis, other than the standard basis $\mathcal{E} = \{\ve_1, \dots, \ve_n\}$.

Set-up. Let $\mathcal{B} = \{\vv_1, \dots, \vv_n\}$ be an ordered basis for a vector space $V$. This means that any $\vx \in V$ can be uniquely written as a linear combination of members of the basis:

\[\vx = \sum_{i=1}^n a_i \vv_i. \]

The scalars $a_1, \dots, a_n$ are called the coordinates of $\vx$ with respect to the basis $\mathcal{B}$. The map $C_\mathcal{B}: V \to \R^n$ sending each vector $\vx \in V$ to its (ordered) tuple of coordinates is clearly a linear isomorphism between $V$ and $\R^n$.

Definition: Let $T: V \to V$ be a linear transformation. The {matrix representation of $T$ with respect to $\mathcal{B}$} is

\[[T]_\B \coloneqq \mqty[\CB(T(\vv_1)) & \cdots & \CB(T(\vv_n))] \in \mathbb{M}_n. \]

Example: Let $\mathcal{P}_2$ be the set of polynomials of degree $\le 2$, with ordered basis $\B = \{1, x, x^2\}$. The derivative operator $D: \mathcal{P}_2 \to \mathcal{P}_2$ is a linear transformation with matrix representation

\[[T]_\B = \mqty[0&1&0\\0&0&2\\0&0&0]. \]

Theorem: The diagram

V \rar{T} \dar[swap]{\CB} & V \dar{\CB}
\R^n \rar{[T]_\B} & \R^n


Proof: It suffices to check commutativity on a basis. Observe that

\[\begin{tikzcd} \vv_j \rar[mapsto]{T} \dar[mapsto,swap]{\CB} & T(\vv_j) \dar[mapsto]{\CB} \\ \ve_j \rar[mapsto]{[T]_\B} & \text{$j$th col.\ of $[T]_\B$} \end{tikzcd} \]

clearly commutes by our definition of $[T]_\B$. QED

Example: Let $W$ be a $k$-dimensional linear subspace of $\R^n$, and let $T: \R^n \to \R^n$ be the orthogonal projection operator $T = \operatorname{proj}_W$ from $\R^n$ onto $W$. Let

\[\B = \{\vv_1, \dots, \vv_k, \vv_{k+1}, \dots, \vv_n\} \]

be an ordered basis of $\R^n$, where $\vv_1, \dots, \vv_k$ is a basis for $W$, and $\vv_{k+1}, \dots, \vv_n$ is a basis for $W^\perp$. Then clearly,

\[T(\vv_i) = \begin{cases} \vv_i & \text{if } 1 \le i \le k \\ \vo &\text{if } k+1 \le i \le n. \end{cases} \]

This means that

\[[T]_\B = \mqty[I_k & 0_{k,n-k} \\ 0_{n-k,k} & 0_{n-k}]. \]

How can we use this to figure out the matrix representation $[T]_\mathcal{E}$ with respect to the standard basis $\mathcal{E}$?

Change-of-Coordinates Matrix. Suppose we have two ordered bases $\B_1, \B_2$ for $V$. The change-of-coordinates matrix $P$ between $\B_1$ and $\B_2$ is defined by $P = \mqty[C_{\B_2} \circ C_{\B_1}^{-1}]$. This is equivalent to defining $P$ so that the diagram

\[\begin{tikzcd} & V \dlar[swap]{C_{\B_1}} \drar{C_{\B_2}} & \\ \R^n \ar{rr}{P} & & \R^n \end{tikzcd} \]

commutes. We write $\B_1 \overset{P}{\rightsquigarrow} \B_2$, and say that $P$ changes $\B_1$-coordinates to $\B_2$-coordinates.

{How to find $\bm{P}$?} Say $\B_1 = \{\vv_1, \dots, \vv_n\}$. Then using the preceding diagram,

\[\begin{tikzcd} & \vv_j \dlar[swap,mapsto]{C_{\B_1}} \drar[mapsto]{C_{\B_2}} & \\ \ve_j \ar[mapsto]{rr}{P} & & \text{$j$th col.\ of $P$} \end{tikzcd} \]

so we conclude that

\[P = \mqty[C_{\B_2}(\vv_1) & \cdots & C_{\B_2}(\vv_n)]. \]

Example: Let $W \subseteq \R^3$ be the plane defined by $x_1 - 2x_2 + x_3 = 0$, and let $T = \operatorname{proj}_W$. Find $[T]_\mathcal{E}$.

Solution: First, find bases for $W$ and $W^\perp$. We will use the basis

\[\B = \qty{\mqty[2\\1\\0], \mqty[-1\\0\\1], \mqty[1\\-2\\1]} \]

and the matrix representation of $T$ with respect to $\mathcal{B}$ is

\[[T]_\B = \mqty[1&0&0\\0&1&0\\0&0&0]. \]

Let $\mathcal{E} \overset{P}{\rightsquigarrow} \mathcal{B}$, where

\[P^{-1} = \mqty[C_\mathcal{E}(\vv_1) & C_\mathcal{E}(\vv_2) & C_\mathcal{E}(\vv_3)] = \mqty[2&-1&1\\1&0&-2\\0&1&1]. \]

(Note that we define $P^{-1}$, not $P$, because we know the coordinates of $\mathcal{B}$ with respect to the standard basis, not the other way around.) Then

\[P = \frac{1}{6}\mqty[ 2 & 2 & 2 \\ -1 & 2 & 5 \\ 1 & -2 & 1] \]

and using the change-of-basis formula, we have

\[[T]_\mathcal{E} = P^{-1} [T]_\mathcal{B}P = \frac{1}{6} \mqty[ 5 & 2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5]. \]

Lecture 35 (2016-04-11)

Eigenvalues and Eigenvectors. Recall from last lecture that if we have two ordered bases $\B_1 = \{\vv_1, \dots, \vv_n\}$ and