Diagonalizable Matrix - Explanation

To see the meaning of the concept of ‘diagonalizable matrix’, let's do the following:

1. use the column vectors of \(\Mtrx{C}\) to form an ordered basis \(\OrdVSpcBss{B}=(\Vect{b}_1,\dots ,\Vect{b}_n)\) of \(\RNrSpc{n}\).
2. Then \(\Mtrx{C}=\CoordTrafoMtrx{C}{S}{B}\) converts from \(\OrdVSpcBss{B}\)-coordinates to standard coordinates.
3. Both matrices \(\Mtrx{A}\) and \(\Mtrx{D}\) represent the same linear transformation of \(\RNrSpc{n}\). The difference is: \(\Mtrx{A}\) describes it in standard coordinates, while \(\Mtrx{D}\) describes it in \(\OrdVSpcBss{B}\)-coordinates.

Now, denote the diagonal entries of \(\Mtrx{D}\) by \(d_1,\dots ,d_n\):

\[ \Mtrx{D} = \left[ \begin{array}{cccc} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & & \vdots \\ \vdots & & \ddots & 0 \\ 0 & \cdots & 0 & d_n \end{array} \right] \]

Then

\(\Mtrx{D}\cdot (\Vect{b}_i)_{\EuScript{B}}\)

\(=\)

\(d_i\cdot \Vect{b}_i\)

In other words, \(\Vect{b}_i\) is an eigenvector of \(\Mtrx{A}\) with eigenvalue \(d_i\).

Conclusion: Given \(\Mtrx{A}\) we may try to diagonalize it by looking for a basis of \(\RNrSpc{n}\) consisting of eigenvectors of \(\Mtrx{A}\).