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Suppose B = ($v_1, ..., v_n$) and C = ($w_1, ... ,w_m$) be ordered bases for V and W respectively. (Finite-dimensional)

I have come to understand that $\varphi$: Hom(V,W) $\to M_{m x n}$, where $\varphi(T) = [T]_{B}^{C}$ is an isomorphism.

I am wondering what a matrix representation of such an isomorphism might look like?

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  • $\begingroup$ Matrix representation with respect to which basis? You should specify this because for any isomorphism, if we are free to choose the bases, then we can always arrange it so that we get the identity matrix (the answer below shows how for this specific case, but it's true now generally as well). $\endgroup$
    – peek-a-boo
    Jul 9 '21 at 16:09
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If $T$ is an element of $\hom(V,W)$ one can prescribe $$Tv_k=\sum_s T_{sk}w_s$$ and then assign $\varphi(T)=[T_{ij}]$.

This map will have

i) $\varphi(T+S)= \varphi(T)+\varphi(S)$,

ii) $\varphi(rT)=r\varphi(T)$ for each scalar $r$.

iii) trivial $\ker(\varphi)$ and

iv) for each matrix in $M_{m\times n}$ a corresponding linear transformation $V\to W$.

But for the matrix of $\varphi$ one constructs the basic linear transformations $V\to W$ (defined at the basis) as: $$L_{rt}v_k=\delta_{rk}w_t$$ which generates $\hom(V,W)$ and then assign $$\varphi(L_{rt})=[\delta^i_r\delta^j_t].$$ Here the matrices $[\delta^i_r\delta^j_t]=:E_{rt}$ have and $1$ at the entry $rt$ and zeroes elsewhere, these generate any element in $M_{m\times n}$.

So the $\varphi$'s matrix is constructed by observing something like that $$L_{11}\mapsto E_{11},$$ $$L_{12}\mapsto E_{12},$$ $$...$$ $$L_{mn}\mapsto E_{mn}.$$

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  • $\begingroup$ here $\delta_{kl}=\delta^k_l$ are the famous Kronecker's delta which give you an $1$ if $k=l$ and zero otherwise. $\endgroup$
    – janmarqz
    Jul 9 '21 at 16:23
  • $\begingroup$ as personal polite habit I always upvote the question that gave me up-scores $\endgroup$
    – janmarqz
    Jul 9 '21 at 16:27
  • $\begingroup$ additional politeness also to my competitors $\endgroup$
    – janmarqz
    Jul 9 '21 at 16:56
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    $\begingroup$ Thank you, I do the same! $\endgroup$
    – Gauss
    Jul 9 '21 at 17:05
  • $\begingroup$ I am glad the effect I was seeking works! Thank you very much @Gauss $\endgroup$
    – janmarqz
    Jul 9 '21 at 17:07
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To do this, first we need two basis, one for the domain and one for the codomain.

Let $T_{ij}: V \to W$ be defined as $T_{ij}(v_i) = w_j$ and $T_{ij}(v_k) = 0$ if $k \neq i$ and let $\mathcal{B} = (T_{11}, ..., T_{mn})$. If $T$ is a linear transformation from $V$ to $W$, we know $T(v_i) = \alpha_{i1} w_1 + ... + \alpha_{im} w_m$, which is equal to $\alpha_{i1}T_{i1}(v_i) + ... + \alpha_{im}T_{im}(v_i) = \alpha_{11}T_{11}(v_i) + ... + \alpha_{nm}T_{nm}(v_i)$, since every term with $k \neq i$ is equal to $0$. Therefore, $\mathcal{B}$ generates the homomorphism space.

They are also Linearly independent: you can check that by applying an arbitrary combination on a $v_i$, which will result in a linear combination of the $w_j$ yielding 0.

The matrix representation of each $T_{ij}$ is a matrix containing $1$ in the $i,j$ position, and zeroes elsewhere.

For the base of the matrices, we consider $\mathcal{C} = (E_{11}, ..., E_{mn})$, where each $E_{ij}$ has a $1$ in the $i,j$ position, and zeroes elsewhere. But wait! Those are the matrix representations of $T_{ij}$!

So, $\varphi(T_{ij}) = E_{ij}$. Using the ordering as I have, the matrix representation of $\varphi$ is precisely the identity matrix, as it takes the $k$-th element of base $\mathcal{B}$ onto the $k$-th element of base $\mathcal{C}$

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Here is a small example of representing the above transformation as a matrix:

enter image description here

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