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Let $V$ a $\mathbb R-$vector space of dimension $m$ and $W$ a $\mathbb R-$vector space of dimension $n$. We denote $$\mathcal L(V,W)=\{\varphi:V\to W\mid \varphi\text{ linear}\},$$ and $$\mathcal M_{n\times m} =\{\text{matrix }m\times n\}.$$

In my course it's written $$\mathcal L(V,W)\cong \mathcal M_{n\times m}(\mathbb R).$$

I'm not sure what is really mean. For me it mean that they are isomorphic, but but since for a linear map $\varphi:V\to W$, the matrix of $\varphi$ is depending on $\varphi$, changing the basis of $V$ and/or $W$ will give an other matrix to $\varphi$, and thus the map wouldn't be injective. So does $$\mathcal L(V,W)\cong \mathcal M_{n\times m}(\mathbb R),$$ really make sense ?


I however totally agree that if $\mathcal B$ is a basis of $V$ and $\mathcal B'$ is a basis of $W$, then

\begin{align*} \mathcal L(V,W)&\longrightarrow \mathcal M_{n\times m}(\mathbb R)\\ \varphi&\longmapsto (\varphi)_{\mathcal B'\mathcal B} \end{align*} is an isomorphism.

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    $\begingroup$ It means they are isomorphic. To show an isomorphism you have to fix a basis, as you said. Hence, it is true they are isomorphic up to a choice of a basis for $V$ and $W$. $\endgroup$
    – Federico
    Jun 23 '18 at 13:42
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The notation $\mathcal A\cong \mathcal B$ mean that "there exist an isomorphism from $A$ to $B$". Since $$\varphi\longmapsto (\varphi)_{\mathcal B'\mathcal B},$$ is an isomorphism between $\mathcal L(V,W)$ and $\mathcal M_{m\times n}(\mathbb R)$, then, indeed $$\mathcal L(V,W)\cong \mathcal M_{m\times n}(\mathbb R).$$

But, as your remarked this isomorphism is not canonical (i.e. it depend on the choice of a basis).

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Both $\mathcal{L}(V,W)$ and $\mathcal{M}_{n\times m}(\mathbb{R})$ are algebras over the reals. Actually, they are isomorphic algebras, and the map that you defined is such an isomorphism.

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Well! It actually makes sense you see, matrix corresponding to any linear transformation in $L(V,W)$ is actually an $m * n$ matrix and for any $m * n $ matrix $A $ we can always define a linear transformation like $T(x)= A x$ Hence there is a one to one correspondence between the desired vector spaces preserving the corresponding operations Therefore they are isomorphic.

Another easier way of understanding is since both spaces are of dimension $m*n$ Hence they will surely be isomorphic. Hope! It works..

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