Show that $f$ is differentiable and find the formula of $Df(I)(H)$.

Question

Consider the function $f:M_{n\times n}(\mathbb R)\to M_{n\times n}(\mathbb R)$ with the formula given below:
$f(A)=A^t+A^2A^t$

($I$ is the identity function.)

So, i was thinking that maybe i could see this as a combination of some functions. If we consider that:
$f_1:X\mapsto X^t$
$f_2:X\mapsto X^2$
$f_3: (X,Y)\mapsto XY$
$f_4: (X,Y)\mapsto X+Y$

Then we have:

$f(A)=f_1(A)+f_3(f_1(A)+f_2(A))$

Can i use this? Can i say:
$Df(A)=Df_1(A)+Df_3(f_1(A)+f_2(A))\times (Df_1(A)+Df_2(A))$
Is this true?

score 1 · Answer 1 · answered May 25 '17 at 14:39

Your starting point is fine, you just have to work on it now:

$$\Bbb d f (I) = \Bbb d (A^t) (I) + \Bbb d (A^2) (I) I^t + I^2 \Bbb d (A^t) (I) = (\Bbb d A)^t (I) + \Bbb d (A^2) (I) + (\Bbb d A)^t (I) = \\ 2 (\Bbb d A)^t (I) + [(\Bbb d A) A + A (\Bbb d A)] (I) = 2 (\Bbb d A)^t (I) + 2 (\Bbb d A) (I) .$$

If $A = (a_{ij})$ then $\Bbb d A = (\Bbb d a_{ij})$, so $(\Bbb d A) (I) = (\Bbb d A)^t (I) = (\Bbb d a_{ij})$, therefore $\Bbb d f (I) = 2 (\Bbb d A)^t + 2 (\Bbb d A)$ and evaluating in $H$ simply amounts to replacing $\Bbb d A$ by $H$, i.e. $\Bbb d f (I) (H) = 2 H^t + 2 H$.

Show that $f$ is differentiable and find the formula of $Df(I)(H)$.

1 Answers1