Diagonalization

Writing a matrix as $P D P^{- 1}$ , where the columns of $P$ are eigenvectors and $D$ holds the eigenvalues. It is possible exactly when a basis of eigenvectors exists, and it turns repeated application of the matrix into arithmetic on $n$ numbers.

Definition

An $n \times n$ matrix $A$ is diagonalizable when there is an invertible $P$ and a diagonal $D$ with

A = P D P^{- 1}, equivalently P^{- 1} A P = D .

What $P$ and $D$ contain. If $A$ has eigenvectors $v_{1}, \dots, v_{n}$ forming a basis, with $A v_{j} = λ_{j} v_{j}$ , take $P = [v_{1} \dots v_{n}]$ and $D = diag (λ_{1}, \dots, λ_{n})$ . Then $A P = P D$ column by column, and $P$ is invertible because its columns are independent. The $j$ th column of $P$ and the $j$ th diagonal entry of $D$ must correspond.

The criterion. $A$ is diagonalizable over a field $F$ exactly when

the characteristic polynomial splits over $F$ , and
for every eigenvalue, geometric multiplicity equals algebraic multiplicity.

Equivalently: the eigenspace dimensions sum to $n$ . A matrix failing (2) is defective and is diagonalizable over no field; a matrix failing only (1) becomes diagonalizable over a splitting field.

Sufficient condition. $n$ distinct eigenvalues force diagonalizability, since each contributes at least one dimension and the algebraic multiplicities are all 1. It is not necessary: $I$ has a single eigenvalue and is already diagonal.

What it is for. Powers become elementwise:

A^{k} = P D^{k} P^{- 1}, D^{k} = diag (λ_{1}^{k}, \dots, λ_{n}^{k}),

because the inner factors cancel in $P D P^{- 1} P D P^{- 1} \dots$ . The same substitution evaluates any polynomial in $A$ , and extends to $e^{A} = P e^{D} P^{- 1}$ , which is what solves a linear system of differential equations.

Similarity. $A$ and $P^{- 1} A P$ are similar, and similar matrices share their characteristic polynomial, eigenvalues with multiplicities, determinant, trace and rank. Diagonalization is the search for the simplest matrix similar to $A$ ; when none is diagonal, the Jordan form is the next-best normal form.

Assumptions and scope

The columns of $P$ and the diagonal entries of $D$ must be in the same order. Permuting the eigenvectors permutes the eigenvalues identically, and mismatching them gives a false factorisation that fails on multiplication.
$P$ is not unique. Scaling any eigenvector, or reordering the basis, gives another valid $P$ , and an eigenspace of dimension above 1 admits infinitely many choices.
Diagonalizability is relative to a field. A real matrix may be diagonalizable over $C$ and not over $R$ , and saying a matrix is not diagonalizable without naming the field is incomplete.
Distinct eigenvalues are sufficient but not necessary. A repeated eigenvalue is compatible with diagonalizability when its eigenspace is large enough.
Symmetric real matrices are always diagonalizable, and by an orthogonal $P$ . That is the spectral theorem and it is a stronger statement than anything provable from the criterion alone.
The factorisation is a tool for hand computation and theory. Numerically, forming $P^{- 1}$ can be ill-conditioned when eigenvectors are nearly dependent, and practical algorithms avoid it.

Worked material

Example

Four verdicts

Distinct eigenvalues: decided without computing an eigenspace. $B = (\begin{matrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 4 & 5 & - 1 \end{matrix})$ is lower triangular, so its eigenvalues are $2$ , $3$ and $- 1$ , three distinct values in a $3 \times 3$ . Each $det (B - λ I) = 0$ . By the corollary it is diagonalizable, and no eigenspace need be computed to know that. Computing them is still required to build $P$ .

Already diagonal. $diag (7, - 2)$ is diagonalizable with $P = I$ and $D$ equal to itself. This is the case a procedure should recognise rather than grind through.

Repeated eigenvalue, still diagonalizable. $S = 3 I = (\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix})$ has the single eigenvalue 3 with algebraic multiplicity 2. Here $S - 3 I$ is the zero matrix, so its kernel is all of $R^{2}$ and the geometric multiplicity is 2 as well. Every nonzero vector is an eigenvector; any basis of $R^{2}$ serves as the columns of $P$ . The sum of geometric multiplicities is 2, so the criterion is met.

This is the case that refutes "repeated root means defective". The repetition is compatible with diagonalizability; the size of the eigenspace decides it.

Repeated eigenvalue, defective. $Q = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$ has the same characteristic polynomial $(λ - 3)^{2}$ as $S$ , same eigenvalue, same algebraic multiplicity. But $Q - 3 I = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ has rank 1, so its kernel is one-dimensional: $E_{3} = span {(1, 0)}$ .

The geometric multiplicities sum to 1, short of 2, so $Q$ is not diagonalizable. Any attempted $P$ would need two columns drawn from a one-dimensional space, making them dependent and $P$ singular, for instance $(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ has determinant 0.

The decisive pair. $S$ and $Q$ have identical characteristic polynomials and opposite verdicts. Nothing visible in the polynomial distinguishes them; the rank of $A - λ I$ does. That is why step 3 of the procedure cannot be skipped whenever an eigenvalue repeats.

Non-example

Two obstructions, and three mistakes

Obstruction 1: the polynomial does not split. $R = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ has $p (λ) = λ^{2} + 1$ , with no real roots. Over $R$ there are no eigenvalues at all, so certainly no basis of eigenvectors, and $R$ is not diagonalizable there.

Over $C$ the roots are $\pm i$ , distinct, so $R$ is diagonalizable with $D = diag (i, - i)$ and columns $(1, - i)$ and $(1, i)$ . The obstruction was the field, and enlarging it removed the obstruction entirely.

Obstruction 2: not enough eigenvectors. $Q = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$ has $p (λ) = (λ - 3)^{2}$ , which splits over $R$ already. There is no larger field to move to. The failure is that $rank (Q - 3 I) = 1$ , so $E_{3}$ is one-dimensional and only one independent eigenvector exists.

Rank is unchanged by field extension, so $Q$ is not diagonalizable over $C$ either, nor over any field containing 3. This obstruction is permanent, and the Jordan form $(\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$ , already in that form, is the best available.

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Mistake: concluding from the repeated root. " $(λ - 3)^{2}$ , so not diagonalizable." $S = 3 I$ has the same polynomial and is diagonal. The root's multiplicity permits a shortfall without implying one; only $\dim E_{λ}$ decides.

Mistake: mismatching $P$ and $D$ . Listing eigenvectors as columns in one order and eigenvalues on the diagonal in another. Each column then satisfies the wrong equation, $A P \neq P D$ , and the error is silent until the product is checked, which is why the check is step 7 rather than optional.

Mistake: writing $A = P^{- 1} D P$ . The correct form is $A = P D P^{- 1}$ with the eigenvectors as the columns of $P$ . The reversed version diagonalises with respect to the wrong basis and gives wrong answers for powers, while looking symmetric enough to pass unexamined.

The distinction to carry. Missing roots are a property of the field and are removable. Missing eigenvectors are a property of the matrix and are not. Reporting "not diagonalizable" without saying which one applies leaves out the part that decides whether anything can be done about it.

Common errors

Common misconception

A matrix with a repeated eigenvalue cannot be diagonalized, because distinct eigenvalues are what make a basis of eigenvectors possible.

Diagonalization

Assumptions and scope

Worked material

Four verdicts

Two obstructions, and three mistakes

Common errors

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