Not every matrix is diagonalizable. This section is about handling matrices that aren't. There are two approaches. The first one is described in the first video. Matrices whose characteristic polynomials have distinct roots are always diagonalizable. This means that ``almost every" matrix is diagonalizable, and a non-diagonalizable matrix can always be perturbed slightly to turn it into a diagonalizable matrix. For instance, $$A= \begin{pmatrix} 1&1 \cr 0&1 \end{pmatrix},$$ is not diagonalizable, but $$A_\epsilon = \begin{pmatrix} 1 & 1 \cr \epsilon^2 & 1 \end{pmatrix}$$ is diagonalizable for all nonzero values of $\epsilon$. The eigenvalues are $1 \pm \epsilon$, and the eigenvectors are $\begin{pmatrix} 1 \cr \pm \epsilon \end{pmatrix}$.

In general, to handle a problem involving a non-diagonalizable matrix $A$, we can

1. Perturb the matrix slightly into a diagonalizable matrix $A_\epsilon$.
2. Diagonalize $A_\epsilon = P_\epsilon D_\epsilon P_\epsilon^{-1}$.
3. Use all of our techniques for diagonalizable matrices to solve the problem with $A_\epsilon$ in place of $A$, and
4. Take a limit as $\epsilon \to 0$. This often results in indeterminate forms, which we can handle with L'Hopital's rule, treating $\epsilon$ as our variable.

A second approach is to use power vectors. While eigenvectors satisfy $(A-\lambda I){\bf x} = 0$, power vectors of degree $p$ satisfy $$(A-\lambda I)^p {\bf x} = 0, \qquad \hbox{ but } (A-\lambda I)^{p-1} {\bf x} \ne 0.$$ Eigenvectors are power vectors of degree 1.

The wonderful thing about power vectors is that, no matter what $n \times n$ matrix $A$ we work with, we can always find a basis of ${\bf C}^n$ consisting of power vectors. If we choose the basis $mathcal{B}$ correctly, then $[A]_\mathcal{B}$ won't necessarily be diagonal, but it will be block diagonal, with each Jordan block taking the form $$ \begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \cr 0 & \lambda & 1 & \cdots & 0 \cr \vdots & \vdots & \ddots & \ddots & 0 \cr 0&0& \cdots & \lambda & 1 \cr 0&0&\cdots & 0 & \lambda \end{pmatrix}. $$

In the third video, we explore how $A^n$ and $e^{At}$ act on power vectors. If ${\bf z}$ is a power vector of degree $p$, then \begin{eqnarray*} A^n {\bf z} & = & (\lambda I + (A-\lambda I))^n {\bf z} \cr &=& \sum_{j=0}^n {n \choose j} \lambda^{n-j} (A - \lambda I)^j {\bf z} \cr & = & \lambda^n {\bf z} + n\lambda^{n-1} (A-\lambda I){\bf z} + \cdots + {n \choose p-1} (A - \lambda I)^{p-1} {\bf z}, \end{eqnarray*} since $(A - \lambda I)^j {\bf z} = 0$ for all $j \ge p$. Similarly, \begin{eqnarray*} e^{At} {\bf z} & = & e^{(\lambda I + (A-\lambda I))t} {\bf z} \cr &=& \sum_{j=0}^\infty \frac{t^j e^{\lambda t}}{j!} (A - \lambda I)^j {\bf z} \cr & = & \sum_{j=0}^{p-1} \frac{t^j e^{\lambda t}}{j!} (A - \lambda I)^j {\bf z}. \end{eqnarray*}