Existence of \(\Mtrx{A}\)   We begin by expressing the vectors \(L(\StdBssVec{j})\) of \(\RNrSpc{m}\) in coordinates

\[L(\StdBssVec{j}) = (a_{1j},\dots ,a_{mj}) = a_{1j}\StdBssVec{1} + \dots + a_{mj}\StdBssVec{m}\]

Now if

\[\Vect{x} = (x_1,\dots ,x_n) = x_1\StdBssVec{1} + \dots + x_n\StdBssVec{n}\]

then the following computation shows that \(L(\Vect{x})\) can be computed from the vectors \(L(\StdBssVec{1}),\dots ,L(\StdBssVec{n})\) alone

\(L(\Vect{x})\)	\(=\)	\(L(x_1\StdBssVec{1} + \cdots + x_n\StdBssVec{n})\)
\(\)	\(= \)	\(x_1 L(\StdBssVec{1}) + \cdots + x_n L(\StdBssVec{n})\)
\(\)	\(= \)	\(x_1(a_{11}\StdBssVec{1} + \cdots + a_{m1}\StdBssVec{m}) + \cdots + x_n(a_{1n}\StdBssVec{1} + \cdots + a_{mn}\StdBssVec{m})\)
\(\)	\(=\)	\( \begin{array}{cccccc} (a_{11}x_1 & + & \cdots & + & a_{1n}x_n)\StdBssVec{1} + \\ \vdots & & & & \vdots \\ (a_{m1}x_1 & + & \cdots & + & a_{mn}x_n)\StdBssVec{m} \end{array} \)
\(\)	\(=\)	\( \left[ \begin{array}{ccc} a_{11} & \dots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \dots & a_{mn} \end{array} \right] \left[ \begin{array}{c} x_1 \\ \vdots \\ x_n \end{array} \right] \)

This means exactly that \(L\) can be computed by the matrix product stated in the theorem.

Uniqueness of \(\Mtrx{A}\)   It remains to show the matrix \(\Mtrx{A}\) we found above is the only matrix with the property \(L(\Vect{x}) = \Mtrx{A}\Vect{x}\). So suppose \(\Mtrx{B}\) is another matrix satisfying

\[\Mtrx{A}\Vect{x} = L(\Vect{x}) = \Mtrx{B}\Vect{x},\quad \text{for all $\Vect{x}\in \RNrSpc{n}$}\]

Choosing \(\Vect{x} = \StdBssVec{j}\), we find

\(j\)-th column of \(\Mtrx{A} = A\StdBssVec{j} = L(\StdBssVec{j}) = B\StdBssVec{j} =\) \(j\)-th column of \(\Mtrx{B}\).

This holds for each \(j\) with \(1\leq j\leq n\), and so \(\Mtrx{A} = \Mtrx{B}\), as was to be shown.