15.2 Diagonalization of nonsymmetric matrices

If a matrix $A$ 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

$$\fbox{$\displaystyle A' = \left( \begin{array}{cccc}... ...vdots & \vdots & \vdots & \ddots \end{array} \right) $}$$ (15.4)

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 $A'$.

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 $A'$ 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

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

since $\vec e_1,\vec e_2,\ldots$ are eigenvectors of $A$. 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 $A'$ above accomplishes.

Recall from section 15.1 that the transformation matrix $P$ for change of basis to the eigenvectors must equal the matrix $E$ of eigenvectors. You therefore have for any vector $\vec v$ and matrix $M$ that you want to transform from new coordinates to old or vice-versa:

$$\fbox{$\displaystyle \vec v = E \vec v^{\rm\prime} \quad... ... \vec v \qquad M = E M' E^{-1} \quad M' = E^{-1} M E $}$$ (15.5)

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 $E$.