Derivative of square of a matrix

Question

In this paper I'm reading, they say that if

$\Psi(A) = A^2$ then the derivative $D\Psi(A)$ is defined by $D\Psi(A)(X) = AX + XA.$

I'm a bit confused by that line since I know the differential is given by $d(A^2) = A\, dA + dA\, A$.

But to me this looks a little different. I don't see how the $X$ would pop up in the derivative $AX+XA$ in place of $dA$ when we go from a differential to a derivative.

Can someone explain where I'm going wrong and clarify?

what would be $dA$ if $A$ is a matrix ? – Surb Sep 08 '20 at 08:19 — Surb, Sep 08 '20 at 08:19

score 3 · Accepted Answer · answered Sep 08 '20 at 08:22

You have for every matrix $H$:

$\Psi(A+H)=(A+H)^{2}=(A+H)(A+H)=A^{2}+AH+HA+H^{2}=\Psi(A)+AH+HA+H^{2}$,

where the expression $AH+HA$ is linear in $H$. Hence:

$\lim_{H\to 0}\dfrac{\Psi(A+H)-\Psi(A)-(AH+HA)}{||H||}=\lim_{H\to 0}\dfrac{H^{2}}{||H||}=0.$

Therefore, $\Psi$ is differentiable at $A$ and $d\Psi_{A}(H)=AH+HA$ for every $H$.

Derivative of square of a matrix

1 Answers1