14 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

\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 $A'$.

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

\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 $A$. 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 13 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:

\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 $E$.