Prove Theorem 5.11. Suppose \(\operatorname{dim} V=m\) and \(\operatorname{dim}
U=n .\) Then \(\operatorname{dim}[\operatorname{Hom}(V, U)]=m n\).
Suppose \(\left\\{v_{1}, \ldots, v_{m}\right\\}\) is a basis of \(V\) and
\(\left\\{u_{1}, \ldots, u_{n}\right\\}\) is a basis of \(U .\) By Theorem \(5.2\),
a linear mapping in Hom \((V, U)\) is uniquely determined by arbitrarily
assigning elements of \(U\) to the basis elements \(v_{i}\) of \(V\). We define
$$
F_{i j} \in \operatorname{Hom}(V, U), \quad i=1, \ldots, m, \quad j=1, \ldots,
n
$$
o be the linear mapping for which \(F_{i j}\left(v_{i}\right)=u_{j}\), and \(F_{i
j}\left(v_{k}\right)=0\) for \(k \neq i\). That is, \(F_{i j}\) maps \(v_{i}\) into
\(u_{j}\) and the ther \(v\) 's into 0 . Observe that \(\left\\{F_{i j}\right\\}\)
contains exactly \(m n\) elements; hence, the theorem is proved if we show hat
it is a basis of \(\operatorname{Hom}(V, U)\).
Proof that \(\left\\{F_{i j}\right\\}\) generates \(\operatorname{Hom}(V, U) .\)
Consider an arbitrary function \(F \in \operatorname{Hom}(V, U) .\) Suppose
$$
w_{k}=a_{k 1} u_{1}+a_{k 2} u_{2}+\cdots+a_{k n} u_{n}, \quad k=1, \ldots, m,
\quad a_{i j} \in K
$$
Consider the linear mapping \(G=\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i j} F_{i j}
.\) Because \(G\) is a linear combination of the \(F_{i j}\), the proof that
\(\left\\{F_{i j}\right\\}\) generates \(\operatorname{Hom}(V, U)\) is complete if
we show that \(F=G\).
We now compute \(G\left(v_{k}\right) ; k=1, \ldots, m .\) Because \(F_{i
j}\left(v_{k}\right)=0\) for \(k \neq i\) and \(F_{k i}\left(v_{k}\right)=u_{i}\),
$$
\begin{aligned}
G\left(v_{k}\right) &=\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i j} F_{i
j}\left(v_{k}\right)=\sum_{j=1}^{n} a_{k j} F_{k
j}\left(v_{k}\right)=\sum_{j=1}^{n} a_{k j} u_{j} \\
&=a_{k 1} u_{1}+a_{k 2} u_{2}+\cdots+a_{k n} u_{n}
\end{aligned}
$$
Thus, by \((1), G\left(v_{k}\right)=w_{k}\) for each \(k .\) But
\(F\left(v_{k}\right)=w_{k}\) for each \(k .\) Accordingly, by Theorem \(5.2, F=G\);
hence, \(\left\\{F_{i j}\right\\}\) generates \(\operatorname{Hom}(V, U)\).
Proof that \(\left\\{F_{i j}\right\\}\) is linearly independent. Suppose, for
scalars \(c_{i j} \in K\),
$$
\sum_{i=1}^{m} \sum_{j=1}^{n} c_{i j} F_{i j}=0
$$
For \(v_{k}, k=1, \ldots, m\)
$$
\begin{aligned}
0=\mathbf{0}\left(v_{k}\right) &=\sum_{i=1}^{m} \sum_{j=1}^{n} c_{i j} F_{i
j}\left(v_{k}\right)=\sum_{j=1}^{n} c_{k j} F_{k
j}\left(v_{k}\right)=\sum_{j=1}^{n} c_{k j} u_{j} \\
&=c_{k !} u_{1}+c_{k 2} u_{2}+\cdots+c_{k n} u_{n}
\end{aligned}
$$
But the \(u_{i}\) are linearly independent; hence, for \(k=1, \ldots, m\), we have
\(c_{k 1}=0 ; c_{k 2}=0, \ldots, c_{k n}=0 .\) In other words, all the \(c_{i
j}=0\), and so \(\left\\{F_{i j}\right\\}\) is linearly independent.
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