6.6: Determinants. Jacobians. Bijective Linear Operators

We assume the reader to be familiar with elements of linear algebra. Thus we only briefly recall some definitions and well-known rules.

Definition

Given a linear operator $\phi : E^{n} \rightarrow E^{n}\left(\text { or } \phi : C^{n} \rightarrow C^{n}\right),$ with matrix
$$[\phi]=\left(v_{i k}\right), \quad i, k=1, \ldots, n,$$
we define the determinant of $[\phi]$ by
$$\begin{aligned} \operatorname{det}[\phi]=\operatorname{det}\left(v_{i k}\right) &=\left|\begin{array}{cccc}{v_{11}} & {v_{12}} & {\dots} & {v_{1 n}} \\ {v_{21}} & {v_{22}} & {\dots} & {v_{2 n}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {v_{n 1}} & {v_{n 2}} & {\dots} & {v_{n n}}\end{array}\right| \\[12pt] &=\sum(-1)^{\lambda} v_{1 k_{1}} v_{2 k_{2}} \ldots v_{n k_{n}} \end{aligned}$$
where the sum is over all ordered $n$-tuples $\left(k_{1}, \ldots, k_{n}\right)$ of distinct integers $k_{j}\left(1 \leq k_{j} \leq n\right),$ and
$$\lambda=\left\{\begin{array}{ll}{0} & {\text { if } \prod_{j<m}\left(k_{m}-k_{j}\right)>0 \text { and }} \\ {1} & {\text { if } \prod_{j<m}\left(k_{m}-k_{j}\right)<0}\end{array}\right.$$

Recall (Problem 12 in §2) that a set $B=\left\{\vec{v}_{1}, \vec{v}_{2}, \ldots, \vec{v}_{n}\right\}$ in a vector space $E$ is a basis iff
(i) $B$ spans $E,$ i.e., each $\vec{v} \in E$ has the form
$$\vec{v}=\sum_{i=1}^{n} a_{i} \vec{v}_{i}$$
for some scalars $a_{i},$ and
(ii) this representation is unique.
The latter is true iff the $\vec{v}_{i}$ are independent, i.e.,
$$\sum_{i=1}^{n} a_{i} \vec{v}_{i}=\overrightarrow{0} \Longleftrightarrow a_{i}=0, i=1, \ldots, n.$$
If $E$ has a basis of $n$ vectors, we call $E$ n-dimensional (e.g., $E^{n}$ and $C^{n} )$.
Determinants and bases satisfy the following rules.
(a) Multiplication rule. If $\phi, g : E^{n} \rightarrow E^{n}\left(\text { or } C^{n} \rightarrow C^{n}\right)$ are linear, then
$$\operatorname{det}[g] \cdot \operatorname{det}[\phi]=\operatorname{det}([g][\phi])=\operatorname{det}[g \circ \phi]$$
(see §2, Theorem 3 and Note 4).
(b) If $\phi(\vec{x})=\vec{x}$ (identity map), then $[\phi]=\left(v_{i k}\right)$, where
$$v_{i k}=\left\{\begin{array}{ll}{0} & {\text { if } i \neq k \text { and }} \\ {1} & {\text { if } i=k}\end{array}\right.$$
hence det $[\phi]=1 .(\text { Why } ?)$ See also the Problems.
(c) An $n$ -dimensional space $E$ is spanned by a set of $n$ vectors iff they are independent. If so, each basis consists of exactly $n$ vectors.

Definition

For any function $f : E^{n} \rightarrow E^{n}$ (or $f : C^{n} \rightarrow C^{n} ),$ we define the $f$-induced Jacobian map $J_{f} : E^{n} \rightarrow E^{1}\left(J_{f} : C^{n} \rightarrow C\right)$ by setting
$$J_{f}(\vec{x})=\operatorname{det}\left(v_{i k}\right),$$
where $v_{i k}=D_{k} f_{i}(\vec{x}), \vec{x} \in E^{n}\left(C^{n}\right),$ and $f=\left(f_{1}, \ldots, f_{n}\right)$.
The determinant
$$J_{f}(\vec{p})=\operatorname{det}\left(D_{k} f_{i}(\vec{p})\right)$$
is called the Jacobian of $f$ at $\vec{p}$.
By our conventions, it is always defined, as are the functions $D_{k} f_{i}$.

Explicitly, $J_{f}(\vec{p})$ is the determinant of the right-side matrix in formula $(14)$ in §3. Briefly,

$$J_{f}=\operatorname{det}\left(D_{k} f_{i}\right).$$

By Definition 2 and Note 2 in §5,

$$J_{f}(\vec{p})=\operatorname{det}\left[d^{1} f(\vec{p} ; \cdot)\right].$$

If $f$ is differentiable at $\vec{p}$,

$$J_{f}(\vec{p})=\operatorname{det}\left[f^{\prime}(\vec{p})\right].$$

Note 1. More generally, given any functions $v_{i k} : E^{\prime} \rightarrow E^{1}(C),$ we can define a map $f : E^{\prime} \rightarrow E^{1}(C)$ by

$$f(\vec{x})=\operatorname{det}\left(v_{i k}(\vec{x})\right);$$

briefly $f=\operatorname{det}\left(v_{i k}\right), i, k=1, \ldots, n$.

We then call $f$ a functional determinant.

If $E^{\prime}=E^{n}\left(C^{n}\right)$ then $f$ is a function of $n$ variables, since $\vec{x}=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$. If all $v_{i k}$ are continuous or differentiable at some $\vec{p} \in E^{\prime},$ so is $f ;$ for by $(1), f$ is a finite sum of functions of the form

$$(-1)^{\lambda} v_{i k_{1}} v_{i k_{2}} \dots v_{i k_{n}},$$

and each of these is continuous or differentiable if the $v_{i k_{i}}$ are (see Problems 7 and 8 in §3).

Note 2. Hence the Jacobian map $J_{f}$ is continuous or differentiable at $\vec{p}$ if all the partially derived functions $D_{k} f_{i}(i, k \leq n)$ are.

If, in addition, $J_{f}(\vec{p}) \neq 0,$ then $J_{f} \neq 0$ on some globe about $\vec{p}.$ (Apply Problem 7 in Chapter 4, §2, to $\left|J_{f}\right|.)$

In classical notation, one writes

$$\frac{\partial\left(f_{1}, \ldots, f_{n}\right)}{\partial\left(x_{1}, \ldots, x_{n}\right)} \text { or } \frac{\partial\left(y_{1}, \ldots, y_{n}\right)}{\partial\left(x_{1}, \ldots, x_{n}\right)}$$

for $J_{f}(\vec{x}) .$ Here $\left(y_{1}, \ldots, y_{n}\right)=f\left(x_{1}, \ldots, x_{n}\right)$.

The remarks made in §4 apply to this "variable" notation too. The chain rule easily yields the following corollary.

Corollary $\PageIndex{1}$

If $f : E^{n} \rightarrow E^{n}$ and $g : E^{n} \rightarrow E^{n}$ (or $f, g : C^{n} \rightarrow C^{n})$ are differentiable at $\vec{p}$ and $\vec{q}=f(\vec{p}),$ respectively, and if

$$h=g \circ f,$$

then

$$J_{h}(\vec{p})=J_{g}(\vec{q}) \cdot J_{f}(\vec{p})=\operatorname{det}\left(z_{i k}\right),$$

where

$$z_{i k}=D_{k} h_{i}(\vec{p}), \quad i, k=1, \ldots, n;$$

or, setting

$$\begin{aligned}\left(u_{1}, \ldots, u_{n}\right) &=g\left(y_{1}, \ldots, y_{n}\right) \text { and } \\\left(y_{1}, \ldots, y_{n}\right) &=f\left(x_{1}, \ldots, x_{n}\right) \text { ("variables")}, \end{aligned}$$

we have

$$\frac{\partial\left(u_{1}, \ldots, u_{n}\right)}{\partial\left(x_{1}, \ldots, x_{n}\right)}=\frac{\partial\left(u_{1}, \ldots, u_{n}\right)}{\partial\left(y_{1}, \ldots, y_{n}\right)} \cdot \frac{\partial\left(y_{1}, \ldots, y_{n}\right)}{\partial\left(x_{1}, \ldots, x_{n}\right)}=\operatorname{det}\left(z_{i k}\right),$$

where

$$z_{i k}=\frac{\partial u_{i}}{\partial x_{k}}, \quad i, k=1, \ldots, n.$$

Proof

By Note 2 in §4,

$$\left[h^{\prime}(\vec{p})\right]=\left[g^{\prime}(\vec{q})\right] \cdot\left[f^{\prime}(\vec{p})\right].$$

Thus by rule (a) above,

$$\operatorname{det}\left[h^{\prime}(\vec{p})\right]=\operatorname{det}\left[g^{\prime}(\vec{q})\right] \cdot \operatorname{det}\left[f^{\prime}(\vec{p})\right],$$

i.e.,

$$J_{h}(\vec{p})=J_{g}(\vec{q}) \cdot J_{f}(\vec{p}).$$

Also, if $\left[h^{\prime}(\vec{p})\right]=\left(z_{i k}\right),$ Definition 2 yields $z_{i k}=D_{k} h_{i}(\vec{p})$.

This proves (i), hence (ii) also. $\quad \square$

In practice, Jacobians mostly occur when a change of variables is made. For instance, in $E^{2},$ we may pass from Cartesian coordinates $(x, y)$ to another system $(u, v)$ such that

$$x=f_{1}(u, v) \text { and } y=f_{2}(u, v).$$

We then set $f=\left(f_{1}, f_{2}\right)$ and obtain $f : E^{2} \rightarrow E^{2}$,

$$J_{f}=\operatorname{det}\left(D_{k} f_{i}\right), \quad k, i=1,2.$$

Example (passage to polar coordinates)

Let $x=f_{1}(r, \theta)=r \cos \theta$ and $y=f_{2}(r, \theta)=r \sin \theta$.

Then using the "variable" notation, we obtain $J_{f}(r, \theta)$ as

$$\begin{aligned} \frac{\partial(x, y)}{\partial(r, \theta)}=\left|\begin{array}{ll}{\frac{\partial x}{\partial r}} & {\frac{\partial x}{\partial \theta}} \\ {\frac{\partial y}{\partial r}} & {\frac{\partial y}{\partial \theta}}\end{array}\right| &=\left|\begin{array}{cc}{\cos \theta} & {-r \sin \theta} \\ {\sin \theta} & {r \cos \theta}\end{array}\right| \\ &=r \cos ^{2} \theta+r \sin ^{2} \theta=r. \end{aligned}$$

Thus here $J_{f}(r, \theta)=r$ for all $r, \theta \in E^{1} ; J_{f}$ is independent of $\theta$.

We now concentrate on one-to-one (invertible) functions.

Theorem $\PageIndex{1}$

For a linear map $\phi : E^{n} \rightarrow E^{n}\left(\text {or} \phi : C^{n} \rightarrow C^{n}\right),$ the following are equivalent:

(i) $\phi$ is one-to-one;

(ii) the column vectors $\vec{v}_{1}, \ldots, \vec{v}_{n}$ of the matrix $[\phi]$ are independent;

(iii) $\phi$ is onto $E^{n}\left(C^{n}\right)$;

(iv) $\operatorname{det}[\phi] \neq 0$.

Proof

Assume (i) and let

$$\sum_{k=1}^{n} c_{k} \vec{v}_{k}=\overrightarrow{0}.$$

To deduce (ii), we must show that all $c_{k}$ vanish.

Now, by Note 3 in §2, $\vec{v}_{k}=\phi\left(\vec{e}_{k}\right);$ so by linearity,

$$\sum_{k=1}^{n} c_{k} \vec{v}_{k}=\overrightarrow{0}$$

implies

$$\phi\left(\sum_{k=1}^{n} c_{k} \vec{e}_{k}\right)=\overrightarrow{0}.$$

As $\phi$ is one-to-one, it can vanish at $\overrightarrow{0}$ only. Thus

$$\sum_{k=1}^{n} c_{k} \vec{e}_{k}=\overrightarrow{0}.$$

Hence by Theorem 2 in Chapter 3, §§1-3, $c_{k}=0, k=1, \ldots, n,$ and (ii) follows.

Next, assume (ii); so, by rule (c) above, $\left\{\vec{v}_{1}, \ldots, \vec{v}_{n}\right\}$ is a basis.

Thus each $\vec{y} \in E^{n}\left(C^{n}\right)$ has the form

$$\vec{y}=\sum_{k=1}^{n} a_{k} \vec{v}_{k}=\sum_{k=1}^{n} a_{k} \phi\left(\vec{e}_{k}\right)=\phi\left(\sum_{k=1}^{n} a_{k} \vec{e}_{k}\right)=\phi(\vec{x}),$$

where

$$\vec{x}=\sum_{k=1}^{n} a_{k} \vec{e}_{k} \text { (uniquely).}$$

Hence (ii) implies both (iii) and (i). (Why?)

Now assume (iii). Then each $\vec{y} \in E^{n}\left(C^{n}\right)$ has the form $\vec{y}=\phi(\vec{x}),$ where

$$\vec{x}=\sum_{k=1}^{n} x_{k} \vec{e}_{k},$$

by Theorem 2 in Chapter 3, §§1-3. Hence again

$$\vec{y}=\sum_{k=1}^{n} x_{k} \phi\left(\vec{e}_{k}\right)=\sum_{k=1}^{n} x_{k} \vec{v}_{k};$$

so the $\vec{v}_{k}$ span all of $E^{n}\left(C^{n}\right).$ By rule (c) above, this implies (ii), hence (i), too. Thus (i), (ii), and (iii) are equivalent.

Also, by rules (a) and (b), we have

$$\operatorname{det}[\phi] \cdot \operatorname{det}\left[\phi^{-1}\right]=\operatorname{det}\left[\phi \circ \phi^{-1}\right]=1$$

if $\phi$ is one-to-one (for $\phi \circ \phi^{-1}$ is the identity map). Hence $\operatorname{det}[\phi] \neq 0$ if (i) holds.

For the converse, suppose $\phi$ is not one-to-one. Then by (ii), the $\vec{v}_{k}$ are not independent. Thus one of them is a linear combination of the others, say,

$$\vec{v}_{1}=\sum_{k=2}^{n} a_{k} \vec{v}_{k}.$$

But by linear algebra (Problem 13(iii)), $\operatorname{det}[\phi]$ does not change if $\vec{v}_{1}$ is replaced by

$$\vec{v}_{1}-\sum_{k=2}^{n} a_{k} \vec{v}_{k}=\overrightarrow{0}.$$

Thus $\operatorname{det}[\phi]=0$ (one column turning to $\overrightarrow{0}).$ This completes the proof. $\quad \square$

Note 3. Maps that are both onto and one-to-one are called bijective. Such is $\phi$ in Theorem 1. This means that the equation

$$\phi(\vec{x})=\vec{y}$$

has a unique solution

$$\vec{x}=\phi^{-1}(\vec{y})$$

for each $\vec{y}.$ Componentwise, by Theorem 1, the equations

$$\sum_{k=1}^{n} x_{k} v_{i k}=y_{i}, \quad i=1, \ldots, n,$$

have a unique solution for the $x_{k}$ iff $\operatorname{det}\left(v_{i k}\right) \neq 0$.

Corollary $\PageIndex{2}$

If $\phi \in L\left(E^{\prime}, E\right)$ is bijective, with $E^{\prime}$ and $E$ complete, then $\phi^{-1} \in L\left(E, E^{\prime}\right).$

Proof for $E=E^{n}\left(C^{n}\right)$

The notation $\phi \in L\left(E^{\prime}, E\right)$ means that $\phi : E^{\prime} \rightarrow E$ is linear and continuous.

As $\phi$ is bijective, $\phi^{-1} : E \rightarrow E^{\prime}$ is linear (Problem 12).

If $E=E^{n}\left(C^{n}\right),$ it is continuous, too (Theorem 2 in §2).

Thus $\phi^{-1} \in L\left(E, E^{\prime}\right). \quad \square$

Note. The case $E=E^{n}\left(C^{n}\right)$ suffices for an undergraduate course. (The beginner is advised to omit the "starred" §8.) Corollary 2 and Theorem 2 below, however, are valid in the general case. So is Theorem 1 in §7.

Theorem $\PageIndex{2}$

Let $E, E^{\prime}$ and $\phi$ be as in Corollary 2. Set

$$\left\|\phi^{-1}\right\|=\frac{1}{\varepsilon}.$$

Then any map $\theta \in L\left(E^{\prime}, E\right)$ with $\|\theta-\phi\|<\varepsilon$ is one-to-one, and $\theta^{-1}$ is uniformly continuous.

Proof

Proof. By Corollary 2, $\phi^{-1} \in L\left(E, E^{\prime}\right),$ so $\left\|\phi^{-1}\right\|$ is defined and $>0$ (for $\phi^{-1}$ is not the zero map, being one-to-one).

Thus we may set

$$\varepsilon=\frac{1}{\left\|\phi^{-1}\right\|}, \quad\left\|\phi^{-1}\right\|=\frac{1}{\varepsilon}.$$

Clearly $\vec{x}=\phi^{-1}(\vec{y})$ if $\vec{y}=\phi(\vec{x}).$ Also,

$$\left|\phi^{-1}(\vec{y})\right| \leq \frac{1}{\varepsilon}|\vec{y}|$$

by Note 5 in §2, Hence

$$|\vec{y}| \geq \varepsilon\left|\phi^{-1}(\vec{y})\right|,$$

i.e.,

$$|\phi(\vec{x})| \geq \varepsilon|\vec{x}|$$

for all $\vec{x} \in E^{\prime}$ and $\vec{y} \in E$.

Now suppose $\phi \in L\left(E^{\prime}, E\right)$ and $\|\theta-\phi\|=\sigma<\varepsilon$.

Obviously, $\theta=\phi-(\phi-\theta),$ and by Note 5 in §2,

$$|(\phi-\theta)(\vec{x})| \leq\|\phi-\theta\||\vec{x}|=\sigma|\vec{x}|.$$

Thus for every $\vec{x} \in E^{\prime}$,

$$\begin{aligned}|\theta(\vec{x})| & \geq|\phi(\vec{x})|-|(\phi-\theta)(\vec{x})| \\ & \geq|\phi(\vec{x})|-\sigma|\vec{x}| \\ & \geq(\varepsilon-\sigma)|\vec{x}| \end{aligned}$$

by (2). Therefore, given $\vec{p} \neq \vec{r}$ in $E^{\prime}$ and setting $\vec{x}=\vec{p}-\vec{r} \neq \overrightarrow{0},$ we obtain

$$|\theta(\vec{p})-\theta(\vec{r})|=|\theta(\vec{p}-\vec{r})|=|\theta(\vec{x})| \geq(\varepsilon-\sigma)|\vec{x}|>0$$

(since $\sigma<\varepsilon )$.

We see that $\vec{p} \neq \vec{r}$ implies $\theta(\vec{p}) \neq \theta(\vec{r});$ so $\theta$ is one-to-one, indeed.

Also, setting $\theta(\vec{x})=\vec{z}$ and $\vec{x}=\theta^{-1}(\vec{z})$ in (3), we get

$$|\vec{z}| \geq(\varepsilon-\sigma)\left|\theta^{-1}(\vec{z})\right|;$$

that is,

$$\left|\theta^{-1}(\vec{z})\right| \leq(\varepsilon-\sigma)^{-1}|\vec{z}|$$

for all $\vec{z}$ in the range of $\theta$ (domain of $\theta^{-1})$.

Thus $\theta^{-1}$ is linearly bounded (by Theorem 1 in §2), hence uniformly continuous, as claimed. $\quad \square$

Corollary $\PageIndex{3}$

If $E^{\prime}=E=E^{n}\left(C^{n}\right)$ in Theorem 2 above, then for given $\phi$ and $\delta>0,$ there always is $\delta^{\prime}>0$ such that

$$\|\theta-\phi\|<\delta^{\prime} \text { implies }\left\|\theta^{-1}-\phi^{-1}\right\|<\delta.$$

In other words, the transformation $\phi \rightarrow \phi^{-1}$ is continuous on $L(E), E=$ $E^{n}\left(C^{n}\right).$

Proof

First, since $E^{\prime}=E=E^{n}\left(C^{n}\right), \theta$ is bijective by Theorem 1(iii), so $\theta^{-1} \in L(E)$.

As before, set $\|\theta-\phi\|=\sigma<\varepsilon$.

By Note 5 in §2, formula (5) above implies that

$$\left\|\theta^{-1}\right\| \leq \frac{1}{\varepsilon-\sigma}.$$

Also,

$$\phi^{-1} \circ(\theta-\phi) \circ \theta^{-1}=\phi^{-1}-\theta^{-1}$$

(see Problem 11).

Hence by Corollary 4 in §2, recalling that $\left\|\phi^{-1}\right\|=1 / \varepsilon,$ we get

$$\left\|\theta^{-1}-\phi^{-1}\right\| \leq\left\|\phi^{-1}\right\| \cdot\|\theta-\phi\| \cdot\left\|\theta^{-1}\right\| \leq \frac{\sigma}{\varepsilon(\varepsilon-\sigma)} \rightarrow 0 \text { as } \sigma \rightarrow 0. \quad \square$$