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Let $L$ be the set of all linear transformations from $V$ to $W$ (finite dimension). Prove that these laws make $L$ into a vector space, and compute its dimension.

I have proved the problem. But after watching this post, I could not understand why $L\simeq W^{m}$.(Here $\dim(V)=m$ and $\dim(W)=n$). I could not comment on that post. That's is why I am posting a different question. Thanks.

(Btw, my solution was the same way as described in the first answer of that link.)

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    $\begingroup$ This is nothing else than your solution, in short form. Imagine $End(V,W)$ as set of $m\times n$-matrices. Then we can just write down a basis of $mn$ matrices, using the matrices $E_{ij}$, having coefficient $1$ at place $(i,j)$, and zeroes otherwise. $\endgroup$ – Dietrich Burde Jan 31 '18 at 10:48
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Let $\{v_1,\ldots,v_m\}$ be a basis of $V$. Consider the map$$\begin{array}{rccc}\Psi\colon&L&\longrightarrow&W^m\\&F&\mapsto&\bigl(F(v_1),\ldots,F(v_m)\bigr).\end{array}$$Then $\psi$ is an isomorphism.

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$L$ may be viewed as the set of all $n\times m$ matrices (in fact there is a 1-1 correspondence between linear maps and matrices). This set of matrices is clearly a vector space as it is closed under matrix addition and subtraction, and scalar multiplication. Its dimension is $nm$, easily seen by noting that the matrices with $1$ in the $ij^{th}$ entry and zeros everywhere else form a basis.

Suppose $\{w_1,\ldots,w_n\}$ is a basis of $W$, then $\{(0,\ldots,w_i,\ldots,0):i=1,\ldots,n; w_i \text{ in any slot}\}$ is a basis of $W^n$ and this has dimension $nm$, so the two are isomorphic as they have the same base field and dimension.

For an explicit isomorphism (although it is not the simplest example) you can map the basis element matrix with $1$ in the $ij^{th}$entry to $(0,\ldots,w_i,\ldots,0)$, where $w_i$ is in the $j^{th}$slot.

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