Differential Equations and Linear Algebra, 4th Edition

6.2 Diagonalization of Matrices

Given an $n\times n$ matrix A, we may ask how many linearly independent eigenvectors the matrix A has. In Section 6.1, we saw several examples (with $n=2$ and $n=3$) in which the $n\times n$ matrix A has n linearly independent eigenvectors—the largest possible number. By contrast, in Example 5 of Section 6.1, we saw that the $2\times 2$ matrix

$$\mathbf{\text{A}}=\left[\begin{array}{rr}2& 3\\ 0& 2\end{array}\right]$$

has the single eigenvalue $\lambda =2$ corresponding to the single eigenvector $\mathbf{\text{v}}={\left[\begin{array}{rr}1& 0\end{array}\right]}^T.$

Something very nice happens when the $n\times n$ matrix A does have n linearly independent eigenvectors. Suppose that the eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ (not necessarily distinct) of A correspond to the n linearly independent eigenvectors ${\mathbf{\text{v}}}_1,{\mathbf{\text{v}}}_2,\dots ,{\mathbf{\text{v}}}_n$, respectively. Let

$$\mathbf{\text{P}}=\left[\begin{array}{ccc}|& |& |\\ {\mathbf{\text{v}}}_1& {\mathbf{\text{v}}}_2& \cdots & {\mathbf{\text{v}}}_n\\ |& |& |\end{array}\right]$$ (1)

be the $n\times n$ matrix having these eigenvectors ...