Diagonalization

For many matrices $A$ , the eigenvalues and eigenvectors assemble into a factorization

A = P D P^{- 1},

with $D$ diagonal. When this is possible, questions that are hard for $A$ become easy for $D$ . The clearest example is computing powers.

Matrix powers from the factorization

Powers of a diagonal matrix are computed entrywise. If $D = [5003]$ , then $D^{k} = [5^{k} 0 0 3^{k}]$ . When $A = P D P^{- 1}$ , the inner factors telescope:

A^{2} = (P D P^{- 1}) (P D P^{- 1}) = P D (P^{- 1} P) D P^{- 1} = P D^{2} P^{- 1},

and by induction $A^{k} = P D^{k} P^{- 1}$ for every $k \geq 1$ . Computing a power of $A$ takes one diagonal power plus two matrix products, no matter how large $k$ is.

The factorization

A = P D P^{- 1}

turns

A^{k}

into

P D^{k} P^{- 1}

; the only work that scales with

k

is raising diagonal entries to the

k

th power.

The Diagonalization Theorem

The proof also gives the procedure. Diagonalizing an $n \times n$ matrix has four steps.

Algorithm:

\textsc{Diagonalize}(A)

— factor

A = P D P^{-1}

if possible

1
find the eigenvalues of A from the characteristic equation
2
for each distinct eigenvalue L do
3
compute a basis for the eigenspace Nul(A - L I)
4
collect all eigenvectors found into columns v_1, ..., v_m
5
if m < n then
6
return "not diagonalizable"
too few independent eigenvectors
7
P <- [ v_1 ... v_n ]
eigenvectors as columns
8
D <- diag(L_1, ..., L_n)
matching eigenvalues, same order as P
9
return (P, D)

The order of the columns of $P$ is free, but the diagonal of $D$ must use the same order: the $i$ th diagonal entry is the eigenvalue of the $i$ th column of $P$ .

Diagonalizing a matrix with a repeated eigenvalue

The repeated eigenvalue $- 2$ still gave a two-dimensional eigenspace, so three independent eigenvectors were available. That does not always happen.

Diagonalization aligns the standard action of

A

(mixing coordinates) with a pure axiswise scaling

D

once the eigenvector basis is used.

A defective matrix

Diagonalizability is not automatic.

A matrix that lacks a full set of independent eigenvectors is called defective.

Sufficient and exact conditions

Distinct eigenvalues guarantee diagonalizability, because eigenvectors for distinct eigenvalues are independent.

This is sufficient but not necessary: the first $3 \times 3$ example above is diagonalizable with only two distinct eigenvalues. The sharp criterion handles repeated eigenvalues by comparing dimensions.

Diagonalizability compares two multiplicities per eigenvalue; the matrix is diagonalizable only when geometric equals algebraic for every eigenvalue.

For a triangular matrix the eigenvalues are read off the diagonal, so diagonalizability of, say,

A = 500 - 8 00 17 - 2

is settled immediately: three distinct diagonal entries $5, 0, - 2$ , hence three distinct eigenvalues, hence diagonalizable by the distinct-eigenvalues criterion.

The two multiplicities

Quantity	Definition	Bound
Algebraic multiplicity of $λ$	its multiplicity as a root of $det (A - λ I)$	between $1$ and $n$
Geometric multiplicity of $λ$	$dim Nul (A - λ I)$	between $1$ and the algebraic multiplicity
Diagonalizable?	yes iff geometric $=$ algebraic for every eigenvalue	and multiplicities sum to $n$

Read as a statement about maps, $A = P D P^{- 1}$ says that $x \mapsto A x$ is a pure scaling along the axes of the eigenvector basis. The precise version for transformations between abstract vector spaces is eigenvectors and linear transformations.¹

Lay, Linear Algebra and Its Applications, 5th ed., §5.3 — Diagonalization: the factorization $A = P D P^{- 1}$ and its use for powers, Theorem 5 (the Diagonalization Theorem), Theorem 6 (distinct eigenvalues), and Theorem 7 (eigenspace dimensions versus multiplicities). ↩