# Explain Theorem in Linear Map

This is from Axler's Linear Algebra Done Right:

3.20 Proposition

If $V$ and $W$ are finite dimensional, then $L(V , W )$ is finite dimensional and

dim $L(V,W)=(\dim V)(\dim W)$.

Proof:

This follows from the equation $\dim \,\operatorname{Mat}(m, n, F) = mn$, 3.18, and 3.19.

where 3.18 states:

Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

and 3.19 states:

Suppose that $(v_1 , \ldots , v_n)$ is a basis of $V$ and $(w_1, \ldots ,w_m)$ is a basis of $W$. Then $M$ is an invertible linear map between $L(V , W )$ and $\operatorname{Mat}(m, n, F)$.

Can someone explain the proof of 3.20 more clearly because I do not really follow.

Thanks

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3.19 says there's an invertible linear map between two particular vector spaces. That's the same thing as saying that the two spaces are isomorphic --- are you OK with that?

OK, so the two spaces are isomorphic. Now 3.18 says they have the same dimension. That is, the dimension of $L(V,W)$ is the same as the dimension of $Mat(m,n,F)$. Still OK?

Now, the first sentence of the proof tells you what the dimension of $Mat(m,n,F)$ is --- it's $mn$. And what are $m$ and $n$? Well, you're told about a basis of $V$ with $n$ elements, and a basis for $W$ with $m$ elements, so $m$ is the dimension of $W$ and $n$ is the dimension of $V$. So, $mn=(\dim V)(\dim W)$.

Now put it all together: dimension of $L(V,W)$ equals dimension of $Mat(m,n,F)$ equals $mn$ equals $(\dim V)(\dim W)$. Got it?

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thanks, need to let the definitions sink in... –  mathnoob Nov 6 '12 at 0:07

You can do this without using matrices. Let $\{v_1,\dots,v_m\}$ be a basis of $V$ and let $\{w_1,\dots,w_n\}$ be a basis of $W$. Define $mn$ linear maps as follows:

$$T_{ij}(v_k)=\cases{w_j & if k=i \\ 0 & otherwise}$$

for $1 \le i \le m$ and $1 \le j \le n$. Now prove that $\{T_{ij}\}$ is a basis of $L(V,W)$.

In matrix land, what you are doing is proving that the matrices $E_{ij}$, which have a 1 in the $(i,j)$ position and 0 everywhere else, form a basis for $\mbox{Mat}(m,n,F)$. For example, if $m=n=2$ then $$E_{1,1} = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}, E_{1,2} = \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}, E_{2,1} = \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}, E_{2,2} = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$$

is a basis of $\mbox{Mat}(2,2,F)$.

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