# Revisited: How is $\phi:{\cal{L}}(V,W)\rightarrow M_{m\times n}(F)$ an isomophism of vector spaces?

I'm told in lecture that if $V,W$ are vector spaces over $F$ and ${\cal{L}}(V,W)$ is the vector space of all linear maps $V\rightarrow W$ and ${\scr{B}}$ and ${\scr{C}}$ are bases for $V$ and $W$ respectively, then $\phi:{\cal{L}}(V,W)\rightarrow M_{m\times n}(F)$ defined by $\phi(T)=[T]^{\scr{C}}_{\scr{B}}$ is an isomorphism of vector spaces, where $T \in {\cal{L}}(V,W)$ and $[T]^{\scr{C}}_{\scr{B}}\in M_{m\times n}(F)$.

Now, I'm trying to understand how this theorem completes the proof for the following theorem:

Let $\{v_1,\dots,v_n\}$ be linearly independent set in a finite dimensional vector space $V$ and let $w_1,\dots,w_n$ be arbitrary vectors in a vector space $W$, then there is a linear map $T:V\rightarrow W$ such that $T(v_1)=w_1, T(v_2)=w_2, \dots$, or simply $T(v_i)=w_i$ for all $i=1,2,\dots,n$.

My professor concludes at the end of the proof for the second theorem that "[b]y theorem, such [a] $T$ exists and is unique." Any help? Furthermore, how does the below outline conclude such a theorem?

Let ${\scr{B}}=\{v_1,\dots,v_n\}$ be a basis for $V$ and ${\scr{C}}=\{u_1,\dots,u_m\}$ a basis for $W$. Then

\begin{eqnarray} w_1 & = & a_{11}u_1+a_{21}u_2+\cdots+a_{m1}u_m & = & T(v_1)\\ w_1 & = & a_{12}u_1+a_{22}u_2+\cdots+a_{m2}u_m & = & T(v_2)\\ \vdots &&&&\vdots\\ w_n & = & a_{1n}u_1+a_{2n}u_2+\cdots+a_{mn}u_m & = & T(v_n), \end{eqnarray}

which can be rewritten as

\begin{eqnarray} \begin{pmatrix} u_1&\cdots&u_m \end{pmatrix} \begin{pmatrix} a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{m1}&\cdots&a_{mn} \end{pmatrix}= \begin{pmatrix} T(v_1)\\ \vdots \\ T(v_n) \end{pmatrix}, \end{eqnarray}

thus such a $T$ exists.

Does this really suffice to prove?

• By "theorem" he is referring to the first theorem in this question. – Trancot Apr 23 '13 at 4:22
• Perhaps there's something really basic (or not) I'm missing, but I can't see how the first theorem, a really important and basic one, can help to prove the second theorem, in itself a very important one, too. The uniqueness and existence of such $\,T\,$ follows without even mentioning matrices, leave alone that isomorphism... – DonAntonio Apr 23 '13 at 11:09
• @DonAntonio See my edit. – Trancot Apr 23 '13 at 22:08
• @DonAntonio, yes, that's not common for me either, maybe the idea was for the student to work with the matrices and not to use the argument "map $v_i$ to $w_i$ and extend by linearity" – Vinicius M. Apr 23 '13 at 22:50

Since every $m\times{n}$ matrix arises as the matrix of a linear transformation from V to W with respect to bases $\mathscr{B}$ and $\mathscr{C}$ by your given $\phi(T)$ we know $\phi$ is surjective. And since a linear transformation is uniquely determined by its images on a basis, $\phi$ is injective. Thus, $\phi$ is a bijection and therefore $\mathcal{L}(V,W)$ is isomorphic to $M_{m\times{n}}(F)$.
That was the isomorphism proof. The second proof does not even need that, because it follows naturally from the definition of linear transformation. The only thing that T could be would be (given scalars $c_j$, $j=1,2,...,n$ ) $$T(c_1v_1+c_2v_2+...c_nv_n) = c_1w_1+c_2w_2+...+c_nw_n$$ This is well defined and thus it is a linear transformation (I'm not putting the well defined proof here....)

• See my most recent edit. – Trancot Apr 23 '13 at 22:11

Pick $v_{n+1},\dots,v_{\dim V}$ such that ${\scr B} = \{v_1,\dots,v_{\dim V}\}$ is a basis of $V$. Now choose a base ${\scr C}$ of $W$. Try to write down a matrix $M$ such that the corresponding linear map via $\phi$ (with respect to the basis $\scr B$ and $\scr C$) takes $v_i$ to $w_i$, for $1\leq i \leq n$. Note that it doesn't matter what the map does to the other $v_i$, $i>n$.

• See my most recent edit. – Trancot Apr 23 '13 at 22:10
• This would be the case where $n = \dim V$, be aware that the dimension of $V$ can be greater than $n$, but you've got the idea. – Vinicius M. Apr 23 '13 at 22:47

(if n$\le$ dim(v) and m$\le$dim(W))

for every matrix as $A\in M_{m*n}(F)$ exist a linear translate from V to W so $\phi$ is surjective on the other hand if $\phi(T)=0$ it means that t every base of V maps to 0 so T is zero translation

and so $\phi$ nonsingular and so injective

so $\phi$ is isomorphism

• See my most recent edit. – Trancot Apr 23 '13 at 22:09