Let $F$ be a field, $V$ and $W$ are vector spaces over the field $F$. The dimension of $V$ is $n$ and the dimension of $W$ is $m$, where $m, n$ are natural numbers. Let $\mathcal{L}$ be a vector space of all linear maps from V to W.

Determine the dimension of $\mathcal{L}$ depending on values $m,n$.

I know there should be a solution using isomorphism between $\mathcal{L}$ an a vector space of $m \times n$ matrices, but I can't prove it.

  • $\begingroup$ Do you know that there is a one-to-one correspondence between linear transformations and $n\times m$ matrices over $F$? $\endgroup$
    – user160738
    Nov 30, 2016 at 19:41
  • $\begingroup$ @user160738 I have heard something about it, but I don't have a deep knowledge of that. How can it be useful? $\endgroup$
    – Juice
    Nov 30, 2016 at 19:43
  • $\begingroup$ Because if so by bijection (one to one correspondence) dimension of $\mathcal{L}$ is equal to dimension of space of $n\times m$ matrices $\endgroup$
    – user160738
    Nov 30, 2016 at 19:44
  • 3
    $\begingroup$ @user160738, it is not enough to have simply a bijection; this would not guarantee that they have the same dimension, (for instance, there is a bijection from $\mathbb{R}$ to $\mathbb{R}^2$ but they do not have the same dimension as vector spaces over $\mathbb{R}$). There must exist a bijective linear transformation, i.e, a linear isomorphism $\endgroup$
    – Arthur
    Nov 30, 2016 at 20:35

1 Answer 1


Here is the outline for the proof:

Let $B=\{ v_1,...,v_n\}$ be a basis for $V$ and $C=\{w_1,..., w_m\}$ be a basis for $W$. We will now try to find a basis for $$\mathscr{L}(V,W) = \{T:V\rightarrow W\ |\ T \ \text{is linear} \}. $$ For each element of $\{1,..., m\} \times \{1,..., n\}$, consider the linear transformation $E^{p,q}$ whose image in the basis $B$ is given by $$E^{p,q}(v_i) = \begin{cases} 0, & \text{if $i\neq q$} \\ w_p, & \text{if $i = q$} \end{cases}$$

All is left to do now is to prove that these linear transformations are linearly independent and that they span $\mathscr{L} (V,W)$, that is that they form a basis for that space. Since the set of these linear transformations has $nm$ elements, it follows that dim$\ \mathscr{L} (V,W) = nm$.

The details are for you to fill in, but the main idea is there. You should be able to prove that there are indeed linear transformations satisfying the given image on the basis and that the set is linearly independent and spans $\mathscr{L} (V,W)$ (if you don't, please let me know).

This proof was taken from Hoffman and Kunze's Linear Algebra book. The idea of the proof is pretty much the same as establishing an isomorphism between $\mathscr{L} (V,W)$ and $M_{m \times n}(F)$ (think about the matricial representation of a linear transformation from $V$ to $W$ with respect to the basis $B$ and $C$), which is "the usual" proof.

  • $\begingroup$ This is a nice proof, but I would rather stick to the "classic" one with isomorphism between $\mathcal{L}(V,W)$ and a vector space of matrices. I am afraid I won't be able to reproduce Your proof. I've got stuck with proving surjectivity of the bijection between $\mathcal{L}(V,W)$ and $M_{m \times n }(F)$. $\endgroup$
    – Juice
    Dec 1, 2016 at 17:16
  • $\begingroup$ Which isomorphism did you pick? $\endgroup$
    – Arthur
    Dec 1, 2016 at 21:50
  • $\begingroup$ I have found the solution. Thank You for Your help :-) $\endgroup$
    – Juice
    Dec 2, 2016 at 14:36

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