Diagonalization and matrix powers

Working with a diagonal matrix is child’s play: its product, inverse, power and determinant are visible at a glance. Diagonalization means transforming a matrix into a diagonal matrix in a suitable basis. We saw in week 12 that the matrices of the same transformation in different bases are similar ($A' = P^{-1}AP$). If we choose the eigenvectors as the basis, $A'$ is diagonal and the eigenvalues stand on the diagonal. This page answers two questions: when is this possible, and what is it good for when it is possible.

Golden rules:

1. An $n \times n$ matrix is diagonalizable $\iff$ it has $n$ independent eigenvectors.
2. $n$ distinct eigenvalues are enough. For a repeated eigenvalue, compute the dimension of the eigenspace.
3. The columns of $P$ are the eigenvectors, the eigenvalues in $D$ in the same order.
4. Check without taking $A^{-1}$: $AP = PD$.

Definition and the main theorem

A square matrix $A$ is diagonalizable if there is an invertible $P$ such that $P^{-1}AP = D$ is diagonal. Equivalently: $A = PDP^{-1}$.

Theorem. If $A$ is $n \times n$: $A$ is diagonalizable $\iff$ $A$ has $n$ linearly independent eigenvectors. In that case

$P = [\mathbf{v}_1 \mid \mathbf{v}_2 \mid \cdots \mid \mathbf{v}_n], \qquad D = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n), \qquad A\mathbf{v}_i = \lambda_i\mathbf{v}_i.$

Why? $AP = [A\mathbf{v}_1 \mid \cdots \mid A\mathbf{v}_n] = [\lambda_1\mathbf{v}_1 \mid \cdots \mid \lambda_n\mathbf{v}_n] = PD$. Since the columns are independent, $P$ is invertible, and $P^{-1}AP = D$.

The recipe

1. Solve the characteristic equation: the eigenvalues and their algebraic multiplicities.
2. Find a basis of $E_\lambda$ for each eigenvalue.
3. Count the vectors in all the bases. If there are $n$, it is diagonalizable; if fewer, it is not.
4. Write $P$ with the eigenvectors and $D$ with the corresponding eigenvalues in the same order.

Since eigenvectors of distinct eigenvalues are always independent, putting together vectors from different eigenspaces in step 3 is safe.

Example 1. $A = \begin{bmatrix} 2 & 3 \\ 2 & 1 \end{bmatrix}$. From the previous page: $(3, 2)$ for $\lambda = 4$, $(1, -1)$ for $\lambda = -1$.

$P = \begin{bmatrix} 3 & 1 \\ 2 & -1 \end{bmatrix}, \quad D = \begin{bmatrix} 4 & 0 \\ 0 & -1 \end{bmatrix}, \quad P^{-1} = \frac15\begin{bmatrix} 1 & 1 \\ 2 & -3 \end{bmatrix}.$

Check: $AP = \begin{bmatrix} 6 + 6 & 2 - 3 \\ 4 + 2 & 2 - 1 \end{bmatrix} = \begin{bmatrix} 12 & -1 \\ 8 & 1 \end{bmatrix}$, $PD = \begin{bmatrix} 12 & -1 \\ 8 & 1 \end{bmatrix}$. Equal.

$P$ is not unique: you can scale the eigenvectors or change their order. If you change the order, the diagonal of $D$ changes in the same way.

Example 2 (a repeated eigenvalue, still diagonalizable). The $A = \begin{bmatrix} 3 & 2 & -2 \\ 2 & 3 & -2 \\ 2 & 2 & -1 \end{bmatrix}$ from the previous page: two independent eigenvectors $(-1, 1, 0)$, $(1, 0, 1)$ for $\lambda = 1$ (algebraic 2); $(1, 1, 1)$ for $\lambda = 3$. In total 3:

$P = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}, \qquad D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}.$

When is it not diagonalizable?

If the geometric multiplicity of an eigenvalue is smaller than its algebraic multiplicity, there are not enough eigenvectors.

Example 3. $A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$. Eigenvalue 2 (algebraic 2). $A - 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $E_2 = \operatorname{span}\{(1, 0)\}$. Only one independent eigenvector: not diagonalizable. A matrix of this kind is called defective.

PITFALL · saying “not diagonalizable” when you see a repeated root. A repeated eigenvalue alone says nothing. Example 2 is diagonalizable, Q9 is not; both have a double root. Don’t decide without computing the dimension of the eigenspace.

Powers: Aᵏ = PDᵏP⁻¹

If $A = PDP^{-1}$ then

$A^2 = PDP^{-1}PDP^{-1} = PD^2P^{-1}, \qquad A^k = PD^kP^{-1}.$

The $P^{-1}P$ in the middle disappear. And $D^k$ is just the $k$th power of the diagonal entries. $A^5$ for Example 1:

D^5    = diag(4^5, (-1)^5) = diag(1024, -1)
P D^5  = [3·1024  1·(-1); 2·1024  (-1)(-1)] = [3072 -1; 2048 1]
A^5    = (P D^5) P⁻¹ = (1/5) [3072 - 2   3072 + 3; 2048 + 2   2048 - 3]
       = (1/5) [3070 3075; 2050 2045] = [614 615; 410 409]

Check: trace $614 + 409 = 1023 = 4^5 + (-1)^5$. A single chain of products instead of multiplying directly five times.

The same idea extends to polynomials: $p(A) = P\,p(D)\,P^{-1}$. Applications such as systems of differential equations and Markov chains rest on this formula.

In the language of linear transformations

If $T(\mathbf{x}) = A\mathbf{x}$ and $A$ is diagonalizable, then in the basis $B$ made of eigenvectors the matrix of $T$ is $D$ (week 12, $A' = P^{-1}AP$). That is, in this basis $T$ multiplies each coordinate separately by a number. Diagonalization is finding the coordinate system that shows the transformation in its simplest form.

Common mistakes