# How do you prove that these vector spaces are isomorphic?

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.

• 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

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.

• 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 S. H. 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 S. H. 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

A more fundamental (and clear) way to prove that two vector spaces are isomorphic is to show that there exists an invertible linear transformation between the vector spaces (this is actually the definition of isomorphism). So, in general, for two arbitrary finite dimensional vector spaces we can state the following

Theorem

Let $V$ and $W$ be finite dimensional vector spaces over a field $F$, with $\mathrm{dim}(V)=n$ and $\mathrm{dim}(W)=m$, and let $\beta$ be an ordered basis (o.b.) of $V$ and $\gamma$ be an ordered basis of $W$, then the transformation $\Phi:\mathcal{L}(V,W)\to \mathcal{M}_{m\times n}$ given by

$$\Phi(T)=\left[ T \right]_{\beta}^{\gamma}$$

is an isomorphism.

Note: $\left[ T \right]_{\beta}^{\gamma}$ is the matrix representation of the linear transformation $T$. Let $\beta=\{v_1,v_2,\dots,v_n\}$ and $\gamma=\{w_1,w_2,\dots,w_m\}$, if $T(v_j)=\sum_{k=1}^m B_{kj} w_k$ then the $j$-th column of $B:=\left[ T \right]_{\beta}^{\gamma}$ (denoted as $B_{\dot \ ,\ j}$) is $(B_{1j},B_{2j},\dots,B_{mj})^T$ (for $j\in \{1,2,\dots,n\}$).

Proof

The path is to show that $\Phi$ is linear and invertible (then it's an isomorphism).

(I) $\Phi$ is linear. Let $U$, $T$ $\in \mathcal{L}(V,W)$ and $c\in F$. Then

\begin{align} \Phi(U+cT) & = \left[ U+cT \right]_{\beta}^{\gamma} \end{align}

But, if $U(v_j)=\sum_{k=1}^m A_{kj} w_k$ then the $j$-th column of $A:=\left[ U \right]_{\beta}^{\gamma}$ (denoted as $A_{\dot \ ,\ j}$) is $(A_{1j},A_{2j},\dots,A_{mj})^T$. So

\begin{align} (U+cT)(v_j) & = U(v_j)+cT(v_j) \\ & = \sum_{k=1}^m A_{kj} w_k+c\sum_{k=1}^m B_{kj} w_k \\ & = \sum_{k=1}^m (A_{kj} +c B_{kj}) w_k \end{align}

Which implies that the $j$-th column of $C:=\left[ U+cT \right]_{\beta}^{\gamma}$ (denoted as $C_{\dot \ ,\ j}$) is

\begin{align} C_{\dot \ ,\ j}&=(A_{1j}+cB_{1j},A_{2j}+cB_{1j},\dots,A_{mj}+cB_{1j})^T \\ &=(A_{1j},A_{2j},\dots,A_{mj})^T+c(B_{1j},B_{2j},\dots,B_{mj})^T \\ &=A_{\dot \ ,\ j}+cB_{\dot \ ,\ j} \end{align}

Since this is valid for $j\in \{1,2,\dots,n\}$, we conclude that $C=A+cB$, or $\left[ U+cT \right]_{\beta}^{\gamma}=\left[ U \right]_{\beta}^{\gamma}+c\left[ T \right]_{\beta}^{\gamma}$, i.e. $\Phi(U+cT)=\Phi(U)+c\Phi(T)$.

(II) $\Phi$ is invertible. To show that this is true, it's enough to show that $\Phi$ is both injective and surjective.

A standard theorem (you can check it in any linear algebra book) says that if two linear transformations have the same matrix representation, then they are the same linear transformation. This implies that $\Phi$ is injective.

On the other hand $\Phi$ is surjective because given $M \in \mathcal{M}_{m\times n}$ (with entries $M_{ij}$), there exists a unique linear transformation $L\in \mathcal{L}(V,W)$ such that $\Phi(L)=\left[ L \right]_{\beta}^{\gamma}=M$. This is the transformation that does the following: $L(v_j)=\sum_{k=1}^m M_{ki}w_k$ (which is the same as the $j$-th column of $M$ and is defined by the entries of $M$).

We conclude that $\Phi$ is an isomorphism. $\blacksquare$

• well yes $M_{n \times m}(k)$ is "just" a representation of elements of $L(k^n,k^m)$ in a given base, for example in the canonical base – reuns Oct 9 '16 at 6:51
• The point of this proof is to avoid talking about a basis of $\mathcal{L}(V,W)$. – Ana S. H. Oct 9 '16 at 13:39
• Is there a name for this theorem? – abuchay Feb 7 '18 at 14:42