Given \(\Vect{x}\in \VSpc{W}\), we need to express it as a linear combination of \(\OrdVSpcBss{C}\). So suppose

\[ \begin{array}{rcclrcc} \CoordVect{x}{B} & = & (x_1,\dots ,x_r) & \quad \text{i.e.} & \Vect{x} & = & x_1 \Vect{b}_1 + \cdots + x_r \Vect{b}_r \\ (\Vect{b}_1)_{\OrdVSpcBss{C}} & = & (a_{11},\dots ,a_{r1}) & \quad \text{i.e.} & \Vect{b}_1 & = & a_{11}\Vect{c}_1 + \cdots + a_{r1}\Vect{c}_r \\ \vdots & & \vdots\quad\vdots & & \vdots & & \vdots \qquad \vdots \\ (\Vect{b}_r)_{\OrdVSpcBss{C}} & = & (a_{1r},\dots ,a_{rr}) & \quad \text{i.e.} & \Vect{b}_r & = & a_{1r}\Vect{c}_1 + \cdots + a_{rr}\Vect{c}_r \\ \end{array} \]

This implies

\(\Vect{x}\)	\(=\)	\(x_1 \Vect{b}_1 + \cdots + x_r \Vect{b}_r\)
\(\)	\(=\)	\(x_1\left( a_{11}\Vect{c}_1 + \cdots + a_{r1}\Vect{c}_r\right)\ + \cdots +\ x_r\left( a_{1r}\Vect{c}_1 + \cdots + a_{rr}\Vect{c}_r\right)\)
\(\)	\(=\)	\(\left(a_{11}x_1 + \cdots + a_{1r}x_r\right)\Vect{c}_1\ + \cdots +\ \left( a_{r1}x_1+\cdots + a_{rr}x_r\right)\Vect{c}_r\)

Writing \(\CoordVect{x}{B}\) and \(\CoordVect{x}{C}\) as column vectors we find

\[ \begin{array}{rcl} \CoordVect{x}{C}\ & = & \begin{bmatrix} a_{11}x_1\ +\ \dots\ +\ a_{1r}x_r \\ \vdots\qquad\qquad\qquad\quad\vdots \\ a_{r1}x_1\ +\ \dots\ +\ a_{rr}x_r \end{bmatrix}\ =\ \begin{bmatrix} a_{11} & \dots & a_{1r} \\ \vdots & & \vdots \\ a_{r1} & \dots & a_{rr} \end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_r \end{bmatrix} \\ & = & \CoordTrafoMtrx{C}{C}{B}\, \CoordVect{x}{B} \end{array} \]

It remains to show that \(\CoordTrafoMtrx{C}{C}{B}\) is the only matrix performing conversion from \(\OrdVSpcBss{B}\)-coordinates to \(\OrdVSpcBss{C}\)-coordinates. So suppose

\[ \Mtrx{D}\ =\ \begin{bmatrix} d_{11} & \dots & d_{1r} \\ \vdots & & \vdots \\ d_{r1} & \dots & d_{rr} \end{bmatrix} \]

also satisfies \(\CoordVect{x}{C} = D \CoordVect{x}{B}\) for all vectors \(\Vect{x}\in \VSpc{W}\). Now, if \(1\leq j\leq r\), we find

\[ \begin{bmatrix}a_{1j} \\ \vdots \\ a_{rj} \end{bmatrix}\ =\ (\mathbf{b}_j)_{\OrdVSpcBss{C}}\ =\ \begin{bmatrix} d_{11} & \dots & d_{1r} \\ \vdots & & \vdots \\ d_{r1} & \dots & d_{rr} \end{bmatrix} \begin{bmatrix}0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{bmatrix}\ =\ \begin{bmatrix}d_{1j} \\ \vdots \\ d_{rj} \end{bmatrix} \]

This means exactly that \(\Mtrx{D}=\CoordTrafoMtrx{C}{C}{B}\) as required. – The proof is complete.