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Let be $k$ a field; $n$, $m$ positive integers. And define $$\mathcal{L}\left(k^{n},k^{m}\right)\equiv\left\{ T:k^{n}\to k^{m}|\,\mathrm{T\, is\, a\, linear\ transformation}\right\} $$

How do you prove that $$\mathcal{L}\left(k^{n},k^{m}\right)\cong M_{m\times n}(k)$$ where the last expression means that the vector spaces are isomorphic.

I'm a undergraduate student taking a first course in Linear Algebra.

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I'm afraid this is way too involved to display it reasonably well in this site. Any decent linear algebra must include this important theorem, which shows both sets are not only isomorphic as vectors spaces but even as algebras over the respective field. – DonAntonio Nov 30 '12 at 3:42
@Don: you don't get algebras if $n \neq m$. The correct statement is that the isomorphism sends composition of linear transformations to matrix multiplication. – Qiaochu Yuan Nov 30 '12 at 3:44
Of course, thanks. – DonAntonio Nov 30 '12 at 3:48
up vote 4 down vote accepted

To be isomorphic as finite dimensional vector spaces, you merely need to have the same dimension. A standard basis for $M_{m\times n}(k)$ should be clear, and has $mn$ elements. A standard basis for the other space consists of all maps $f_{ij}$, where $f_{ij}(\vec{e}_k)=\delta_{jk}\vec{e}_i$. So there is again a basis of $mn$ elements.

If $m=n$, then each vector space is actually an algebra, because you can `multiply' vectors together and the result is a vector of the same type. With $n\times n$ matrices, the multiplication is matrix multiplication. With the linear functions space, composition is the multiplication. In this case the two algebras are indeed isomorphic, but a simple vector space isomorphism is not sufficient. It would need to be one such that $\phi(ab)=\phi(a)\phi(b)$. But the vector space isomorphism hinted at in the first paragraph (sending $f_{ij}$ to $1_{ij}$) will have this feature.

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OK, I'm looking for a more illustrative proof, but can you please explain a little more your last statement? The one with $f_{ij}$. – Ana Nov 30 '12 at 3:49
Is it really the case that all you need to show is isomorphism of vector spaces? If $V$ has basis $\{v_1,\ldots,v_k\}$ and $W$ has basis $\{w_1,\ldots,w_k\}$, then you may define an isomorphism $\phi$ from $V$ to $W$ by $\phi(v_i)=w_i$ and extending linearly. – alex.jordan Nov 30 '12 at 3:55
In your case, the matrices with all zeros and a $1$ at the $i,j$ position form a basis for the vector space of matrices, and the functions described above form a basis for the vectors space of linear mappings. (If you want to prove these things, I'm not sure to what degree you'd like to prove them.) Each basis has the same number of elements, so the isomorphism in my last comment may be defined. – alex.jordan Nov 30 '12 at 3:57
Well, I don't understand how do you calculate the dimension of $\mathcal{L}\left(k^{n},k^{m}\right)$. Actually I'm looking for a specific (or explicit) isomorphism between $\mathcal{L}\left(k^{n},k^{m}\right)$ and $M_{m\times n}(k)$, can you help me? – Ana Nov 30 '12 at 4:24
How do you compute the dimension of any vector space? You demonstrate a basis for it, and then count up how many vectors are in that basis. I have shown you what the basis is: $\{f_{1,1},\ldots,f_{m,n}\}$. Each $f_{ij}$ is a well-defined linear map from $k^n$ to $k^m$. You now need to (1) show that these functions are linearly independent (show that $\sum c_{ij}f_{ij}=0_{\text{function}}\implies c_{ij}=0$ for all $i,j$), and (2) that they span the whole space of maps (that any map from $k^n$ to $k^m$ equals some $\sum c_{ij}f_{ij}$ for the right values of $c_{ij}$). – alex.jordan Nov 30 '12 at 4:33

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