# Is it true that $\mathcal L(V,W)\cong \mathcal M_{n\times m}(\mathbb R)$?

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.

• 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 $W$. Jun 23 '18 at 13:42

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 $$\mathcal L(V,W)\cong \mathcal M_{m\times n}(\mathbb R).$$
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.
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..