Let \(L\from \RNrSpc{n} \to \RNrSpc{n}\) denote the linear map represented by \(\Mtrx{A}\) in standard coordinates. Consider now the situation where \(\Mtrx{A}\) is diagonalizable with diagonalizing matrix

\(\Mtrx{C}\)

\(=\)

\([ \Vect{b}_1\ \dots\ \Vect{b}_n ]\)

satisfying

\[ \left[ \begin{array}{cccc} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & & \vdots \\ \vdots &&\ddots & 0 \\ 0 & \cdots & 0 & d_n \end{array} \right]\ =\ \Mtrx{C}^{-1}\, \Mtrx{A}\, \Mtrx{C} \]

We show first that \(\EuScript{B} \DefEq (\Vect{b}_1,\dots , \Vect{b}_n)\) is an ordered basis of \(\RNrSpc{n}\) consisting of eigenvectors of \(\Mtrx{A}\). Indeed, they form a basis of \(\RNrSpc{n}\), because the matrix \(\Mtrx{C}\) is invertible. To see that each vector \(\Vect{b}_i\) is an eigenvector of \(\Mtrx{A}\), we compute:

\[ \begin{array}{rcl} \Mtrx{A} \Vect{b}_i & = & \Mtrx{C} \Mtrx{D} \Mtrx{C}^{-1} \Vect{b}_i \\ & = & \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}} \Mtrx{D} \CoordTrafoMtrx{C}{\EuScript{B}}{\EuScript{S}} \Vect{b}_i \\ & = & \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}}\, \left[ \begin{array}{cccc} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & & \vdots \\ \vdots &&\ddots & 0 \\ 0 & \cdots & 0 & d_n \end{array} \right]\, \left[ \begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{array} \right] \\ & = & \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}}\, \left[ \begin{array}{c} 0 \\ \vdots \\ 0 \\ d_i \\ 0 \\ \vdots \\ 0 \end{array} \right] \\ & = & d_i \Vect{b}_i \end{array} \]

So \(\EuScript{B}\) consists of eigenvectors of \(\Mtrx{A}\), and the eigenvalue of \(\Vect{b}_i\) is \(d_i\). To see how these eigenvalues are related to the characteristic polynomial of \(\Mtrx{A}\), we compute

\(\det(\Mtrx{A} - \lambda \IdMtrx{n})\)	\(=\)	\(\det(\Mtrx{C} \Mtrx{D} \Mtrx{C}^{-1}\ -\ \lambda \IdMtrx{n})\)
\(\)	\(= \)	\(\det(\Mtrx{C} \Mtrx{D} \Mtrx{C}^{-1}\ -\ C(\lambda \IdMtrx{n})C^{-1})\)
\(\)	\(= \)	\(\det(\Mtrx{C} (\Mtrx{D} - \lambda \IdMtrx{n}) \Mtrx{C}^{-1} )\)
\(\)	\(= \)	\(\det( \Mtrx{D} - \lambda \IdMtrx{n} )\)
\(\)	\(=\)	\((\lambda - d_1)\cdots (\lambda - d_n)\)
\(\)	\(= \)	\((\lambda - d_1)^{a_1}\, \cdots\, (\lambda - d_r)^{a_r}\)

Now we turn to the relationship between algebraic and geometric multiplicity of \(d_i\). From the computation above we see that \(d_i\) occurred in exactly \(a_i\) diagonal positions of \(\Mtrx{D}\). Corresponding to each of these diagonal positions there is exactly one vector in the ordered list \(\Vect{b}_1,\dots, \Vect{b}_n\) of basis vectors in \(\EuScript{B}\). It follows that the eigenspace \(E_i\) of \(d_i\) has dimension \(a_i\).

Conversely, assume that conditions 1) and 2) of the theorem are satisfied. By hypothesis, \(\RNrSpc{n}\) has an ordered basis \(\EuScript{B} = (\Vect{b}_1,\dots ,\Vect{b}_n)\) of eigenvectors of \(\Mtrx{A}\). Let \(\lambda_1,\dots ,\lambda_n\) denote the corresponding eigenvalues. We need to show that \(\Mtrx{A}\) is diagonalizable. Let \(\Mtrx{C} \DefEq [\Vect{b}_1\ \dots\ \Vect{b}_n]\) and

\[ \Mtrx{D}\ \DefEq\ \left[ \begin{array}{cccc} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & & \vdots \\ \vdots & & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_n \end{array} \right] \]

Then \(\Mtrx{D}\) represents \(L\) with respect to \(\EuScript{B}\) because, for \(1\leq i\leq n\),

\[L(\Vect{b}_i) = \lambda_i \Vect{b}_i= \Mtrx{D}(\Vect{b}_i)_{\EuScript{B}}\]

Therefore,

\[\Mtrx{D} = \CoordTrafoMtrx{C}{\EuScript{B}}{\EuScript{S}}\, \Mtrx{A}\, \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}} = \Mtrx{C}^{-1} \Mtrx{A}\, \Mtrx{C}\]

and this means exactly that \(\Mtrx{C}\) is diagonalizable, as claimed.