Dimension of Hom(U,V)

Question

I know this has been asked before - I am really struggling to understand what people have said though, so I want to ask for myself.

If U,V are vector spaces over field K, with dimensions n,m respectively then why is the dimension of Hom(U,V) (over k) = mn?

Any help would be appreciated!

Answer 1

$\mathrm{Hom}(U,V)$ is isomorphic to $\mathrm{Mat}(n,m)$, the set of matrices $n \times m$, where $m = \dim U$ and $n = \dim V$. An isomorphism consists of sending each $T \in \mathrm{Hom}(U,V)$ to the matrix $[T]_{\mathcal{C}}^{\mathcal{B}}$, where $\mathcal{B}$ and $\mathcal{C}$ are any fixed bases of $U$ and $V$, respectively. Since isomorphic spaces have the same dimension, $\dim \mathrm{Hom}(U,V) = \dim \mathrm{Mat}(n,m)$. On the other hand, a basis of $\mathrm{Mat}(n,m)$ consists of matrices with one entry being 1 and all the other entries being 0. There are exactly $mn$ such matrices, because $mn$ is the number of entries of a matrix with $n$ rows and $m$ columns. Therefore, $\dim \mathrm{Hom}(U,V) = \dim \mathrm{Mat}(n,m) = mn$.

Answer 2

This is because, it $(e_1, \dots, e_n)$ is a basis for $U$, $(f_1,\dots,f_m)$ is a basis for $F$, $U=\bigoplus\limits_{i=1}^n Ke_i$, $V=\smash{\bigoplus\limits_{j=1}^m} Kf_j$ and $\operatorname{Hom}$ commutes with finite direct sums: $$\DeclareMathOperator\Hom{Hom}\Hom(U,V)\simeq\Hom\Bigl(\bigoplus_{i=1}^n Ke_i,\bigoplus_{j=1}^m Kf_j\Bigr)\simeq \smash{\bigoplus_{1\le i\le n\atop 1\le j\le m}}\Hom(Ke_i,Kf_j)\simeq\smash{\bigoplus_{1\le i\le n\atop 1\le j\le m}}K=K^{mn} $$

Answer 3

It's probably worth adding that in case $U$ has infinite dimension $m$, the space $\text{Hom}(U, V)$ has dimension given by $|V|^m$ where $|V|$ is the cardinality of $V$. For if $B$ is a basis of $U$, each linear map $f: U \to V$ determines and is determined by a function $g: B \to V$, i.e., there is a linear isomorphism

$$\text{Hom}(U, V) \cong V^B$$

and the dimension of $V^B$ equals the cardinality of $V^B$ by this MathOverflow answer: https://mathoverflow.net/questions/49551/dimension-of-infinite-product-of-vector-spaces. (Probably the OP intended that $m, n$ are finite, but the actual question didn't say this and the result is still of interest.)