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# 6.6: Determinants. Jacobians. Bijective Linear Operators

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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

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