Diagonalization

Diagonalizing a matrix is simply rewriting it in a basis made of its eigenvectors, where all it does is stretch each axis by its own eigenvalue. The name comes from the result: in that basis the matrix is diagonal, all zeros except down the main diagonal. That's useful because diagonal matrices are about as easy as matrices get, especially when you need to apply the same one over and over. All we're really saying in $A = PDP^{-1}$ is three steps, read right to left: $P^{-1}$ (translate into eigenvector coordinates), $D$ (stretch each coordinate by its eigenvalue) and $P$ (translate back). It's three simple moves in a row; bear with it.

Why it works. This is the $P^{-1}AP$ from the change-of-basis lesson, with $P$'s columns chosen to be eigenvectors. In the eigenvector basis, $A$ just scales each coordinate by its eigenvalue, and that's $D$. So $P^{-1}$ converts into eigenvector coordinates, $D$ scales, and $P$ converts back.

Powers: notice what happens when you square it: $A^2 = PDP^{-1}PDP^{-1}$, and the $P^{-1}P$ in the middle cancels, leaving $PD^2P^{-1}$. The same thing happens for any power, so $A^k = PD^kP^{-1}$. And $D^k$ just raises each diagonal entry to the $k$th power. This predicts the long-run behavior of repeated processes.

When is $A$ diagonalizable?

• $n$ distinct eigenvalues: always.
• Repeated eigenvalues: only if each eigenspace's dimension equals the eigenvalue's multiplicity. A shear like $\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$ fails.

A common mistake. It's tempting to think a repeated eigenvalue automatically rules out diagonalizing, because the shear fails. It isn't: the identity matrix has $\lambda = 1$ twice and is already diagonal. What matters is whether you can find $n$ independent eigenvectors, not whether the eigenvalues repeat. And when you build $D$, keep the eigenvalues in the same order as their eigenvectors in $P$.

Why it matters: seeing far ahead in one step. A robot's tracker predicts its state one step at a time, $\vec x_{k+1} = A\vec x_k$. To know where it will be in $100$ steps, you could multiply by $A$ a hundred times, which is slow and piles up rounding error. Or diagonalize once: $A^{100} = PD^{100}P^{-1}$, and $D^{100}$ only asks for two numbers raised to the $100$th power. Split the state along the eigenvectors and the whole future is readable at a glance: each piece just gets multiplied by its own eigenvalue every step. There's an animation further down that shows it: the piece with eigenvalue $0.5$ dies out, so the robot's path lines up with the eigenvector whose eigenvalue is $1.1$.

Fibonacci numbers, population projections and Markov chains are computed with $A^k = PD^kP^{-1}$; the largest eigenvalue gives the long-run growth rate.