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There are two finitely dimension vector spaces $V$ and $W$. Dimensions are $n$ and $m$ respectively.

$$L(V,W)=\{T:V\rightarrow W \;|\; T \;\text{is linear}\}$$

$L(V,W)$ is a vector space with operations addition and scalar multiplication.

My approach was to use the basis of $V$ and $W$.

let $\{v_1,\cdots,v_n\}$ be a basis for $V$, then any $v\in V$ can be written as $$a_1 v_1+\cdots+a_n v_n$$

for every $T(v)$ an element of $L(V,W)$ then

$$T(v) = T(a_1 v_1+...+a_n v_n) = a_1T(v_1)+...+a_nT(v_n)$$

let $\{w_1,\cdots,w_m\}$ be a basis for $W$ then every w an element of $W$ can be expressed as $$b_1w_1+...+b_mw_m$$

let $w_i'=T(v_i)$ then

$$T(v)=a_1w'_1+...+a_nw'_n$$

since $w'_i$ is an element of $im(T)$ which is a subspace of $W$ then it can be expressed according to the basis of $W$.

I am not sure if this interpretation is correct, so I seek your feedback.

Ehsan M. Kermani
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Trena
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3 Answers3

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can you see that the given space is nothing but the space of $m\times n$ real matrices?As you fix a basis for $V$ and $W$, each linear transformation is expressible as an $m\times n$ matrix acting on the right of row vectors of length $n$.Now define a natural isomorphism between your space and the space of $m\times n$ real matrices

Arpit Kansal
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If you fix the basis of $V$ and $W$, there exists a unique $m\times n$ matrix of scalars corresponding to a linear map from $V$ to $W$. Show that the space of $m\times n$ real matrices has dimension $mn$. It is quite easy, can you find the basis?

QED
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choose bases $\{v_i\}^{n}_{i=1},\{w_j\}^{m}_{j=1}$ for $V$ and $W$ respectively and now consider $$f_{ij}(v_k) = \begin{cases} w_j &\mbox{if } k=i \\ 0 & \text{otherwise} \end{cases} $$ Now try to show that this forms a basis for $L(V,W)$

Mathronaut
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