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Let $f: \mathbb{R}^{n^{2}} \rightarrow \mathbb{R}^{n^{2}}$ defined by $f(X)=X^6$, where $\mathbb{R}^{n^{2}}$ is the space of square $n\times n$ matrices.

Show that $f$ is a differential application.

I try to prove that using the definition of a differential application in this way:

Let $X \in \mathbb{R}^{n^{2}}$ and $Y\neq0 \in \mathbb{R}^{n^{2}}$.

$$f(X+Y) - f(X) = f'(X)\cdot Y + r(Y) $$ $$\Leftrightarrow r(Y)=f(X+Y) - f(X) - f'(X)\cdot Y $$ $$\Leftrightarrow \frac{r(Y)}{|Y|}=\frac{f(X+Y) - f(X) - f'(X)\cdot Y}{|Y|}.$$ Then $$\lim_{|Y|\rightarrow0}\frac{r(Y)}{|Y|}=\lim_{|Y|\rightarrow0}\frac{f(X+Y) - f(X) - f'(X)\cdot Y}{|Y|} $$ $$\Leftrightarrow\lim_{|Y|\rightarrow0}\frac{r(Y)}{|Y|}=\lim_{|Y|\rightarrow0}\frac{(X+Y)^6 - X^6 - f'(X)\cdot Y}{|Y|}.$$

I want to prove that $\lim_{|Y|\rightarrow0}\frac{r(Y)}{|Y|}=0$. If I could get this, the result follows by the differential application definition. The problem is that the algebraic manipulations aren't so simple, and I don't have any additional properties for these matrices.

I try something else: There is a result that says every function $f:\mathbb{R^n}\rightarrow \mathbb{R^n}$ such that $f \in C^1$, then $f$ is differentiable. I thought to calculate the partial differentials of $f$ in each variable $x_{i} \in \mathbb{R}$ and proof that derivative is continuous in each $x_i$. But I don't know how to calculate these derivatives because they are matrices. Could someone help me, please?

Benjamin Wang
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Pedro Souza
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1 Answers1

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Let $X,H\in\mathbb{R}^{n^2}$, then

\begin{align} f'(X)[H]&=\frac{\partial}{\partial t}\left\{f(X+tH) \right\}_{t=0}\\ &=\frac{\partial}{\partial t}\left\{(X+tH)^{6}\right\}_{t=0}\\ &=6X^{5}H \end{align} Now, \begin{equation} f(X+H)-f(X)=f'(X)[H]+r(H)\Rightarrow r(H)=(X+H)^{6}-X^6-6X^{5}H \end{equation}

We have to \begin{align} 0\leq\frac{\Vert r(H)\Vert}{\Vert H\Vert}=\frac{\Vert 15X^{4}H^{2}+20X^{3}H^{3}+15X^{2}H^{4}+6XH^{5}+H^6\Vert}{\Vert H\Vert}\leq\frac{\Vert 15X^{4}H+20X^{3}H^{2}+15X^{2}H^{3}+6XH^{4}+H^{5}\Vert\Vert H\Vert}{\Vert H\Vert}\quad\text{(See note)} \end{align} Therefore \begin{equation} \lim_{H\to 0}\frac{r(H)}{\Vert H\Vert}=0 \end{equation}

This shows that $f$ is differentiable at $X\in\mathbb{R}^{n^2}$.

Note: Let $T\in L(\mathbb{R}^{n},\mathbb{R}^{m})$ and $K=\{x\in \mathbb{R}^{n}:\Vert x\Vert_{n}=1\}$, then $\Vert T(x)\Vert_{n}\leq \Vert T\Vert \Vert x\Vert_{n}$, with that you can verify that the norm operator is submultiplicative.

logarithm
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