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Lets say that I have the following function: $$ y = (f \circ g \circ h)(x) = f(g(h(x))) $$

$$ f:\mathbb{R}^{k} → \mathbb{R}, g : \mathbb{R}^{m} \to \mathbb{R}^k, h: \mathbb{R}^{n} \to \mathbb{R}^m $$

what is the dimension of the Jacobian matrix $D(f \circ g \circ h)(x)$?

Would the dimension be $1\times k$, since the function is only in function of one variable $x$?

Alex Provost
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Chen
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  • Let $f\circ g\circ h=F:\mathbb{R}^n\to\mathbb{R}$ (a scalar field.) What do you know about the Jacobian of scalar fields? – K.defaoite Apr 29 '21 at 14:52

2 Answers2

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By the Chain Rule you have $$D(f\circ g\circ h)=Df\cdot Dg\cdot Dh,$$ where these last derivatives have:

$Df$ a $1\times k$-matrix;

$Dg$ a $k\times m$-matrix

and

$Dh$ an $m\times n$-matrix

which can be multiplied, then $D(f\circ g\circ h)$ is a $1\times n$-matrix.

That multiplication is of matrices which have functions in their entries.

Further, the evaluated version is $$D(f\circ g\circ h)|_p=Df|_{(g\circ h)(p)}\cdot Dg|_{h(p)}\cdot Dh|_p,$$ which is a product of matrices with numbers as entries, however the same analysis of the number of rows and columns applies, as above.

janmarqz
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The composition $f\circ g\circ h$ is such that $$\underbrace{\mathbb R^n\overset{h}\to\mathbb R^m\overset{g}\to\mathbb R^k\overset f\to\mathbb R}_{\tilde f:\mathbb R^n\to\mathbb R}$$ so this shows that the Jacobian $J$ of the map $\tilde f:=f\circ g\circ h:\mathbb R^n\to\mathbb R$ is an element of $\operatorname{M}(1\times n,\mathbb R)$.

If (assuming a certain regularity of the function), for example, you consider smooth functions, you can think the Jacobian evaluated in a point $p\in\mathbb R^n$ as the representative matrix of the differential (linear) map, so in our case $$\tilde f_{p,*}:T_p\mathbb R^n\to T_{\tilde f(p)}\mathbb R$$ respect to the standard basis of the tangent spaces $T_p\mathbb R^n$ and $T_{\tilde f(p)}\mathbb R$, so respectively $\bigg\{ \dfrac{\partial}{\partial x^i}\bigg |_{p}\bigg \}$ and $\dfrac{d}{dt}\bigg|_{\tilde f(p)}$. The Jacobian can be computed with the chain rule: $$J_{\tilde f}(p)=\underbrace{J_f(g(h(p)))}_{\in\operatorname{M}(1\times k,\mathbb R)}\cdot \underbrace{J_g(h(p))}_{\in\operatorname{M}(k\times m,\mathbb R)}\cdot \underbrace{J_h(p)}_{\in\operatorname{M}(m\times n,\mathbb R)}$$

Vajra
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