Alberto Calhoun

2022-11-18

Let $f:M\left(n,\mathbb{R}\right)\to M\left(n,\mathbb{R}\right)$ and let $f\left(A\right)=A{A}^{t}$. Then find derivative of $f$, denoted by $df$ .
So, Derivative of $f\left(df\right)$ if exists, will satisfy $limH\to 0\frac{||f\left(A+H\right)-f\left(A\right)-df\left(H\right)||}{||H||}=0$.

### Answer & Explanation

Envetenib8ne

We can try calculating $f\left(A+H\right)-f\left(A\right)$ and see what is left
$f\left(A+H\right)-f\left(A\right)=\left(A+H\right)\left(A+H{\right)}^{t}-A{A}^{t}=A{A}^{t}+A{H}^{t}+H{A}^{t}+H{H}^{t}-A{A}^{t}\phantom{\rule{0ex}{0ex}}=A{H}^{t}+H{A}^{t}+H{H}^{t}$
$f\left(A+H\right)-f\left(A\right)=\left(A+H\right)\left(A+H{\right)}^{t}-A{A}^{t}=A{A}^{t}+A{H}^{t}+H{A}^{t}+H{H}^{t}-A{A}^{t}\phantom{\rule{0ex}{0ex}}=A{H}^{t}+H{A}^{t}+H{H}^{t}$
The term $H{H}^{t}$ has norm $‖H{H}^{t}‖=‖H{‖}^{2}$ and so it will go to zero even when divided by $‖H‖$. Thus our candidate is
$d{f}_{A}\left(H\right)=A{H}^{t}+H{A}^{t}$
Now you can try and prove the limit is $0$ explicitly.

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