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I wish to prove that the dimension for the set of all linear mappings $L:V \to W$ is equal to $\dim V \times \dim W$. I know that any general linear mapping can be represented as a matrix, so intuitevely it makes sense that the dimension should be $\dim V \times \dim W$ but I cannot find a way to prove this mathematically

Bernard
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Skrrrrrtttt
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  • Try constructing an isomorphism from $\mathcal L(V,W)\to M_{m\times n}(\mathcal F),$ given that $\dim(V)=n, \dim(W)=m$. – linear_combinatori_probabi Sep 21 '18 at 19:26
  • In general, if $V$ is finite-dimensional (regardless of whether $W$ is finite-dimensional), the equality $$\dim\big(\text{Hom}(V,W)\big)=\dim(V),\dim(W)$$ holds always. However, the equality is false when $V$ is infinite-dimensional. – Batominovski Sep 21 '18 at 19:42
  • @IsanaYashiro would the linear mapping T that takes a linear mapping from L and maps it to its standard matrix work as an isomorphism? – Skrrrrrtttt Sep 21 '18 at 23:17
  • @Skrrrrrtttt: Yes, since $\Phi_{\beta\to\gamma}(T):=[T]_\beta^\gamma$ is isomorphism with any choices of $\beta,\gamma$. – linear_combinatori_probabi Sep 22 '18 at 04:19

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Let $m=\dim V, n=\dim W$. Any linear map $$L:\Bbb R^m\rightarrow\Bbb R^n$$ may be represented as an $n\times m$-matrix $$\mathbf{L}=\begin{pmatrix}a_{1,1}&a_{1,2}&\dots&a_{1,m}\\a_{2,1}&a_{2,2}&\dots&a_{2,m}\\\vdots&\vdots&&\vdots\\a_{n,1}&a_{n,2}&\dots&a_{n,m}\end{pmatrix}$$

How many different such $\mathbf{L}$ can you make?

cansomeonehelpmeout
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