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U and V are linear spaces contained in the same field F

What is the easiest way to prove the following:

dimHom(U,V)=dimU⋅dimV

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    Well it follows since $\text{Hom}(U,V)\cong U^*\otimes V$ and $\dim U^*=\dim U$ if $U$ is finite-dimensional but presumably you don't want to use tensor products. – 2017-02-16

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Let $\{a_i\}$ be a basis for $U$ and $\{b_j\}$ a basis for $V$ consider the maps $f_{ij}$ defined by $$f_{ij}(a_i)=b_j$$ and $$f_{ij}(a_l)=0 \ \ \text{for $l\neq i$}$$

Now show that $\{f_{ij}\}$ is a basis for $Hom(U,V)$.

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    Thank you. Then to show the basis, I would use the tensor product or there would be something more convenient? – 2017-02-16
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    well you could just do it directly if $g(a_i)=\sum \alpha_{ij}b_j$ then $g=\sum \alpha_{ij}f_{ij}$ – 2017-02-16
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    Oh, I see it. Thank you! – 2017-02-16