Let $V,W$ be finite dimensional vector spaces over the same field $K$. Let $n=\dim V$ and $m=\dim W$, we know that $\dim \mathcal{L}(V;W)=nm$. I was thinking about a nice way to prove it, and I came out with the following idea.

We know that every $T \in \mathcal{L}(V;W)$ is completely defined by it's values on the basis of $V$. So consider $\{v_1,\dots,v_n\}$ basis of $V$ and consider $\varphi : \oplus_{i=1}^nW\to\mathcal{L}(V;W)$ a function that given the element $(w_1,\dots,w_n)\in \oplus_{i=1}^n$ gives us back $T \in \mathcal{L}(V;W)$ such that $T(v_i)=w_i$, so we know that $\varphi$ is linear, because:


$$\varphi(\lambda w_1,\dots \lambda w_n)(v_i)=\lambda w_i = \lambda \varphi(w_1,\dots,w_n)(v_i)$$

It is also injective, because if $\varphi(w_1,\dots,w_n)=0$ this means that $\varphi(w_1,\dots,w_n)(v_i)=0$ for every $i$ and hence $w_i=0$ implying that $(w_1,\dots,w_n)=(0,\dots,0)$ and so $\ker \varphi = \{0\}$. In that case, since $V$ and $W$ are finite dimensional, $\varphi$ is a linear isomorphism. Since we know that the dimension of the external direct sum is the sum of the dimensions and since a linear isomorphism preserves dimensions, it follows that $\dim \mathcal{L}(V;W)= \sum_{i=1}^n m = nm$.

Is this proof ok or there is any problem with it?

Thanks very much in advance!


You cannot argue that $\varphi$ is surjective using that it is injective and that the spaces are finite-dimensional; with that point of view you only prove that the dimension of the codomain is bigger than the dimension of the domain.

Now, it is very very easy to show that $\varphi$ is surjective: given any $T\in\mathcal L(V;W)$, define $w_j=T(v_j)$, and then $T=\varphi(w_1,\ldots,w_n)$.

Once you know that $\varphi$ is bijective, you are right in saying that the dimensions agree.

  • $\begingroup$ Yes, you are right about surjectivity. But it is really simple to prove it. Thanks for the help @MartinArgerami! $\endgroup$ – user1620696 Aug 22 '13 at 2:11

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