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Let \(\displaystyle{n\in\mathbb{N}\,,n\geq 2}\). Define a norm in the \(\displaystyle{\mathbb{R}}\) - space \(\displaystyle{\left(\mathbb{M}_{n}\,(\mathbb{R}),+,\cdot\right)}\) such that the \(\displaystyle{\mathbb{R}}\) - spaces \(\displaystyle{\left(\mathbb{B}\,(\mathbb{R}^{n},\mathbb{R}^{n}),+,\cdot\right)}\) and \(\displaystyle{\left(\mathbb{M}_{n}\,(\mathbb{R}),+,\cdot\right)}\) be isometrically isomorphic and then find a base for \(\displaystyle{\left(\mathbb{B}\,(\mathbb{R}^{n},\mathbb{R}^{n}),+,\cdot\right)}\).



Note : \(\displaystyle{\left(\mathbb{R}^{n},||\cdot||\right)=\left(\mathbb{R}^{n},||\cdot||_{2}\right)}\), where \(\displaystyle{||\cdot||_{2}}\) is the \(\displaystyle{\rm{Eucleidean}}\) norm.

Let \(\displaystyle{f\in\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right)}\) . Since \(\displaystyle{f}\) is linear, there exists

\(\displaystyle{A=\left(a_{ij}\right)\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\) such that \(\displaystyle{f(x)=f_{A}(x)=A\cdot x^{T}\,,x\in\mathbb{R}^{n}}\).

If \(\displaystyle{x=\left(x_1,x_2,...,x_n\right)\in\mathbb{R}^{n}}\), then :

\(\displaystyle{f(x)=\left(a_{11}\,x_1+a_{12}\,x_2+...+a_{1n}\,x_n,...,a_{n1}\,x_1+a_{n2}\,x_2+...+a_{nn}\,x_n\right)}\),

so :

\(\displaystyle{||f(x)||_{2}=\sqrt{\sum_{i=1}^{n}\,\left(\sum_{j=1}^{n}a_{ij}\,x_{j}\right)^2}}\) .

Now, by \(\displaystyle{\rm{Cauchy-Schwarz}}\) inequality in the \(\displaystyle{\rm{Hilbert}}\) space

\(\displaystyle{\left(\mathbb{R},+,\cdot,<,>\right)}\), we get :

\(\displaystyle{\forall\,i\in\left\{1,2,...,n\right\}: \left(\sum_{j=1}^{n}a_{ij}\,x_{j}\right)^2\leq \left(\sum_{j=1}^{n}a_{ij}^2\right)\,\left(\sum_{j=1}^{n}x_{j}^2\right)=||x||^2\,\left(\sum_{j=1}^{n}a_{ij}^2\right)}\)

and then :

\(\displaystyle{||f(x)||_{2}\leq \sqrt{\sum_{i=1}^{n}\,\sum_{j=1}^{n}a_{ij}^2\,||x||^2}=\sqrt{\sum_{i=1}^{n}\,\sum_{j=1}^{n}a_{ij}^2}\,||x||}\), so :

\(\displaystyle{||f(x)||_{2}\leq a\,||x||\,,\forall\,x\in\mathbb{R}^{n}}\), where :

\(\displaystyle{a=\sqrt{\sum_{i=1}^{n}\,\sum_{j=1}^{n}a_{ij}^2}\geq 0}\) .

Also, by this progress, we proved that if \(\displaystyle{f=f_{A}}\) is a linear map for some \(\displaystyle{A\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\), then

\(\displaystyle{f}\) is continuous.

Note : \(\displaystyle{||f||=\sup\,\left\{||f(x)||_{2}: x\in\mathbb{R}^{n}\,,||x||_{2}\leq 1\right\}}\).

Therefore,

\(\displaystyle{\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right)=\left\{f_{A}:\mathbb{R}^{n}\longrightarrow \mathbb{R}^{n}: A\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)\right\}}\) .

We define \(\displaystyle{g:\mathbb{M}_{n}\,\left(\mathbb{R}\right)\longrightarrow \mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right)}\) by

\(\displaystyle{g(A)=f_{A}}\).

If \(\displaystyle{A\,,B\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\) and \(\displaystyle{c\in\mathbb{R}}\), then,

for each \(\displaystyle{x\in\mathbb{R}^{n}}\) holds :

\(\displaystyle{\begin{aligned} g(A+B)(x)&=f_{A+B}(x)\\&=\left(A+B\right)\cdot x^{T}\\&=A\cdot x^{T}+B\cdot x^{T}\\&=f_{A}(x)+f_{B}(x)\\&=\left(f_{A}+f_{B}\right)\cdot x^{T}\\&=\left(g(A)+g(B)\right)(x)\end{aligned}}\)

and

\(\displaystyle{\begin{aligned} g(c\,A)(x)&=f_{c\,A}(x)\\&=\left(c\,A\right)\cdot x^{T}\\&=c\,\left(A\cdot x^{T}\right)\\&=c\,f_{A}(x)\\&=(c\,g(A))(x)\end{aligned}}\)

so : \(\displaystyle{g(A+B)=g(A)+g(B)\,,g(c\,A)=c\,g(A)}\), which means that the function \(\displaystyle{g}\) is \(\displaystyle{\mathbb{R}}\) - linear .

Let \(\displaystyle{f\in\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right)}\). Then, \(\displaystyle{f=f_{A}}\) for some

\(\displaystyle{A\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\) and \(\displaystyle{g(A)=f_{A}=f}\).

Therefore, \(\displaystyle{g}\) is onto \(\displaystyle{\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right)}\) .

Consider now \(\displaystyle{A\,,B\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\) such that \(\displaystyle{g(A)=g(B)}\).

So, \(\displaystyle{f_{A}(x)=f_{B}(x)\iff A\cdot x^{T}=B\cdot x^{T}\,,\forall\,x\in\mathbb{R}^{n}}\) .

Setting \(\displaystyle{x=e_{i}\in\mathbb{R}^{n}\,,1\leq i\leq n}\) where \(\displaystyle{\left\{e_{i}: 1\leq i\leq n\right\}}\) is the usual base of

\(\displaystyle{\left(\mathbb{R}^{n},+,\cdot\right)}\), we get : \(\displaystyle{A=B}\), which means that the function \(\displaystyle{g}\) is one-one.

So, \(\displaystyle{\left(\mathbb{M}_{n}\,\left(\mathbb{R}\right),+,\cdot\right)\simeq \left(\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right),+,\cdot\right)}\)

and thus: \(\displaystyle{\dim_{\mathbb{R}}\,\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right)=\dim_{\mathbb{R}}\mathbb{M}_{n}\,\left(\mathbb{R}\right)=n^2}\).

If \(\displaystyle{\left\{E_{ij}\in\mathbb{M}_{n}\,\left(\mathbb{R}\right): 1\leq i\,,j\leq n\right\}}\) is the usual base of \(\displaystyle{\left(\mathbb{M}_{n}\,\left(\mathbb{R}\right),+,\cdot\right)}\), where :

\(\displaystyle{\left(E_{ij}\right)_{ab}=\begin{cases}
1\,\,\,\,,\left(a,b\right)=\left(i,j\right)\\
0\,\,\,\,,\left(a,b\right)\neq \left(i,j\right)
\end{cases}}\)

then, \(\displaystyle{\left\{f_{E_{ij}}\in\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right): 1\leq i\,,j\leq n\right\}}\) is a base for the

\(\displaystyle{\mathbb{R}}\) - space \(\displaystyle{\left(\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right),+,\cdot\right)}\)

We define \(\displaystyle{||\cdot||:\mathbb{R}^{n}\longrightarrow \mathbb{R}\,,A=\left(a_{ij}\right)\mapsto \sqrt{\sum_{i=1}^{n}\,\sum_{j=1}^{n}a_{ij}^2}}\)

and then we have that \(\displaystyle{||\cdot||}\) is a norm at \(\displaystyle{\left(\mathbb{M}_{n}\,\left(\mathbb{R}\right),+,\cdot\right)}\) .

Since \(\displaystyle{\left(\mathbb{B}\,\left(\mathbb{R}^{n},\mathbb{R}^{n}\right),||\cdot||\right)}\) is a \(\displaystyle{\rm{Banach}}\) space, then so

is \(\displaystyle{\left(\mathbb{M}_{n}\,\left(\mathbb{R}\right),||\cdot||\right)}\).

Note : If \(\displaystyle{A=\left(a_{ij}\right)\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\), then :

\(\displaystyle{||A||=\sqrt{\sum_{i=1}^{n}\,\sum_{j=1}^{n}a_{ij}^2}=\sqrt{Tr\,(A\cdot A^{T})}}\) .

Additional question :

Prove that the series \(\displaystyle{\sum_{k=0}^{\infty}\dfrac{A^{k}}{k!}}\) converges for every

\(\displaystyle{A\in\mathbb{M}_{n}\,\left(\mathbb{R}\right)}\) according to the above norm.

I got carried away by this topic and did a little research of my own.. and found some amazing results.

Since the space \( E:=\mathcal M_n(\mathbb C) \) of all \(n×n\) complex matrices is a finite-dimensional vector space, all norms define the same topology. So we can take a sub-multiplicative norm, that is, a norm \( \lVert\cdot\rVert \) such that \( \lVert AB\rVert\leq \lVert A\rVert \cdot\lVert B\rVert \). (For example, we can take \( \lVert\cdot\rVert \) to be the operator norm on E.) As a finite dimensional vector space, E is complete, so it's enough to show normal convergence. We have that, for each integer \(n \geq 0 \) ,
$$0\leq ||A^{n}/n!||\leq \frac{\lVert A\rVert^n}{n!}$$

and we know that, for each real number \( x \) , the series \( \displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} \) converges (it defines the exponential function). Therefore, for any \( A \in E \), the series \( \displaystyle \sum_{n=0}^{\infty}\frac{A^n}{n!} \) converges. (We also got the additional result that \( \displaystyle \lVert e^A\rVert\leq e^{\lVert A\rVert} , \;\; \forall A \in E \).)
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Some comments:
1. If \(A\) is a \(n \times n\) matrix then the series \( \displaystyle \sum_{n=0}^{\infty}\frac{A^n}{n!} \) converges absolutely.
2.If \(A, B\) are two commutative matrices then \(e^{AB} =e^{BA} \) otherwise the result does not necessarily hold true.
3.If \(A, B \) are two commutative matrices then \( \exp (A+B) =\exp A \cdot \exp B =\exp B \cdot \exp A \). This can be proved by Taylor's expansion.. (left as an exercise)