14 Diagonalization of nonsymmetric matrices

If a matrix is not defective, you can use its eigenvectors as new basis. It turns out that in that basis the matrix simplifies to a diagonal matrix

$\begin{displaymath} \fbox{$\displaystyle A' = \left( \begin{array}{cccc}... ...vdots & \vdots & \vdots & \ddots \end{array} \right) $} \end{displaymath}$

(8)

Since this diagonal matrix has the eigenvalues on the main diagonal, (in the order that you arranged the corresponding eigenvectors), it is often written as $\Lambda$ instead of

Needless to say, this simplification is a tremendous help if you are doing analytical or numerical work involving the matrix.

The quickest was to see why is diagonal like above is to note that in terms of the new basis, $A\vec v$ produces a new vector $\vec w^{ \prime}$ according to

$\begin{displaymath} \vec w^{ \prime} = A'\vec v^{ \prime} = A' (v_1' \vec ... ... v_2' \lambda_2 \vec e_2 + \ldots + v_n' \lambda_n \vec e_n \end{displaymath}$

since $\vec e_1,\vec e_2,\ldots$ are eigenvectors of

. So the coefficients of $\vec w^{ \prime}$ are related to those of $\vec w^{ \prime}$ as $v_1'\lambda_1,v_2' \lambda_2,\ldots,v_n' \lambda_n$ . And that is just what multiplying by the diagonal matrix

above accomplishes.

Recall from section 13 that the transformation matrix for change of basis to the eigenvectors must equal the matrix of eigenvectors. You therefore have for any vector $\vec v$ and matrix that you want to transform from new coordinates to old or vice-versa:

$\begin{displaymath} \fbox{$\displaystyle \vec v = E \vec v^{\rm\prime} \quad... ... \vec v \qquad M = E M' E^{-1} \quad M' = E^{-1} M E $} \end{displaymath}$

(9)

Here the primes mean the vector or matrix as it appears in the new basis of eigenvectors.

In summary, a nondefective matrix becomes diagonal when its eigenvectors are used as basis. The main diagonal contains the eigenvalues, ordered like the corresponding eigenvectors in .