Exercise 3.A.3.

Question

Suppose $T \in \mathcal{L}(F^n,F^m)$. Show that there exist scalars $A_{j,k} \in F$ for $j = 1,...,m$ and $k = 1,...,n$ such that
$$
T(x_1,...,x_n) = (A_{1,1}x_1 + \dotsb + A_{1,n}x_n,...,A_{m,1}x_1 + \dotsb + A_{m,n}x_n)
$$
for every $(x_1,...,x_n) = F^n$.
[The exercise above shows that T has the form promised in the last item
of Example 3.4.]

Arden Shi · Accepted Answer

By the homogeneity and additivity of linear maps
\begin{equation}
\begin{split}
T(x_1,...,x_n) &= T(x_1, 0,..., 0) + \dotsb + T(0,...,0,x_n) \
&=x_1 T(1, 0,...,0) + \dotsb + x_n T(0,...,0,1)
\end{split}
\end{equation}
$\exists A_{j,k} \in F$ for $j = 1,...,m$ and $k = 1,...,n$ such that
\begin{equation}
\begin{split}
T(1,0,...,0) &= (A_{1,1},...,A_{m,1}) \
&\cdot \
&\cdot \
&\cdot \
T(0,...0,1) &= (A_{1,n},...,A_{m,n})
\end{split}
\end{equation}
Thus, expanding and combining,
$$
T(x_1,...,x_n) = (A_{1,1}x_1 + \dotsb + A_{1,n}x_n,...,A_{m,1}x_1 + \dotsb + A_{m,n}x_n)
$$

Exercise 3.A.3.

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Exercise 3.A.3.

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