# The Unapologetic Mathematician

## The Jacobian of a Composition

Let’s start today by introducing some notation for the Jacobian determinant which we introduced yesterday. We’ll write the Jacobian determinant of a differentiable function $f$ at a point $x$ as $J_f(x)=\det(df(x))$. Or, in more of a Leibnizean style:

$\displaystyle\frac{\partial(f^1,\dots,f^n)}{\partial(x^1,\dots,x^n)}=\det\left(\frac{\partial f^i}{\partial x^j}\right)$

We’re interested in determining the Jacobian of the composite of two differentiable functions. To which end, suppose $g:X\rightarrow\mathbb{R}^n$ and $f:Y\rightarrow{R}^n$ are differentiable functions on two open regions $X$ and $Y$ in $\mathbb{R}^n$, with $g(X)\subseteq Y$, and let $h=f\circ g:X\rightarrow\mathbb{R}^n$ be their composite. Then the chain rule tells us that

$\displaystyle dh(x)=df(g(x))dg(x)$

where each differential is an $n\times n$ matrix, and the right-hand side is a matrix multiplication.

But these matrices are exactly the Jacobian matrices of the functions! And since the by definition, the determinant of the product of two matrices is the product of their determinants. That is, we find the equation

$\displaystyle J_h(x)=J_f(g(x))J_g(x)$

Or, we could define $y^i=g^i(x)$ and use the Leibniz notation to write

$\displaystyle\frac{\partial(h^1,\dots,h^n)}{\partial(x^1,\dots,x^n)}=\frac{\partial(h^1,\dots,h^n)}{\partial(y^1,\dots,y^n)}\frac{\partial(y^1,\dots,y^n)}{\partial(x^1,\dots,x^n)}$

As a special case, let’s assume that the differentiable function $f:X\rightarrow\mathbb{R}^n$ is injective in some open neighborhood $A$ of a point $a$. That is, every $x\in A$ is sent to a distinct point by $f$, making up the whole image $f(A)$. Further, let’s suppose that the function $f^{-1}$ which sends each point $y\in f(A)$ back to the point in $A$ from which it came — $f^{-1}(y)=x$ if and only if $y=f(x)$ — is also differentiable. Then we have the composition $f^{-1}(f(x))=x$, and thus we find

$\displaystyle J_{f^{-1}}(f(a))J_f(a)=1$

or

$\displaystyle\frac{\partial(y^1,\dots,y^n)}{\partial(x^1,\dots,x^n)}\frac{\partial(x^1,\dots,x^n)}{\partial(y^1,\dots,y^n)}=1$

Thus, if a differentiable function $f$ has a differentiable inverse function defined in some neighborhood of a point $a$, then the Jacobian determinant of the function must be nonzero at that point. A fair bit of work will now be put to turning this statement around. That is, we seek to show that if the Jacobian determinant $J_f(a)\neq0$, then $f$ has a differentiable inverse in some neighborhood of $a$.

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Posted by John Armstrong | Analysis, Calculus

## 3 Comments	»

1. [...] in with a first step towards proving the inverse function theorem we talked about at the end of yesterday’s post. This is going to get [...]

Pingback by | November 13, 2009 | Reply

2. [...] Inverse Function Theorem At last we come to the theorem that I promised. Let be continuously differentiable on an open region , and . If the Jacobian determinant at some [...]

Pingback by | November 18, 2009 | Reply

3. [...] But now we recognize the product of the two Jacobian determinants as the Jacobian of the composition: [...]

Pingback by | January 7, 2010 | Reply

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