Let $M_n(\mathbb{R})$ denotes the space of all $n \times n$ matrices identified with the Euclidean space $\mathbb{R^{n^2}}$. Fix a column vector $x \neq 0$. Define $f:M_n(\mathbb{R}) \rightarrow\mathbb{R}$ by \begin{equation} f(A)= \langle A^2x,x\rangle \end{equation} I am trying to prove $f$ is differentiable.

It looks like you have to find the derivative w.r.t the matrix $A$. How do I do that? Any help is appreciated.

  • $\begingroup$ Just to add another way to think about it. You can show that $f(A)$ is a polynomial in $a_{ij}$. So all $\frac{\partial f}{\partial a_{ij}}$ exist and are continuous, hence, the function has a total derivative by this theorem ( calculus.subwiki.org/wiki/… ) $\endgroup$ – user25959 Dec 7 '18 at 2:12

For bilinear continuous functions $B,$ we have the product rule $$B'(x,y) \cdot (h, k) = B(x,k) + B(h,y).$$

Consider now the bilinear continuous function $\theta(u, v) = (u \mid v)$ (standard inner product in $\mathbf{R}^2$), the bilinear function $\varphi(A_1, A_2) = A_1A_2$ (matrix product), the function $\psi(N) = (Nx, x),$ whose derivative is $\psi'(N) \cdot H = (Hx, 0)$ and the function $\tau(A) = (A, A),$ which is linear, hence $\tau'(A) = \tau.$ The chain rule gives the derivative of $f = \theta \circ \psi \circ \varphi \circ \tau$ (that is, $f$ is differentiable).

Do you want me to edit and spell-out the derivative?


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