Recipe

Test whether a matrix is diagonalizable

Use when: you have a matrix's eigenvalues, at least one of which repeats, and need to know whether n independent eigenvectors actually exist rather than just assuming they do

Goal: a confirmed yes-or-no on diagonalizability, with the exact eigenvalue and shortfall pinpointed if the answer is no

Taught in Similarity, diagonalizability, and defective matrices

1. 1

   Solve the characteristic equation and note how many times each root repeats — that count is its algebraic multiplicity.

   Example

   For A = \begin{bmatrix} 3 & 1 \\ -1 & 5 \end{bmatrix}: \operatorname{tr}(A)=8, \det(A) = 15-(-1) = 16, so p_A(\lambda) = \lambda^2-8\lambda+16 = (\lambda-4)^2 — a single eigenvalue, \lambda=4, with algebraic multiplicity 2.

   • every eigenvalue is distinct (algebraic multiplicity 1 each) — each automatically supplies exactly one independent eigenvector; skip straight to Step 4.
   • some eigenvalue repeats — its eigenspace needs to be checked before anything can be concluded; continue to Step 2.

   Why: an eigenvalue with algebraic multiplicity 1 is guaranteed a 1-dimensional eigenspace (never 0, never more), so only a repeated eigenvalue can possibly cause a shortfall.

2. 2

   Solve (A-\lambda I)\mathbf v = \mathbf 0 and count the free variables — that count is the eigenspace's dimension, the geometric multiplicity.

   Example

   A - 4I = \begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix}. Row 2 is a copy of row 1, leaving one equation -x+y=0 and one free variable, so \mathbf v = t\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \text{geometric multiplicity} = 1

   Why: this is a direct measurement, not a guess — the eigenspace's actual dimension is the only thing that decides whether enough independent eigenvectors exist for this eigenvalue.

3. 3

   Take the geometric multiplicity just measured in Step 2 and compare it to that same eigenvalue's algebraic multiplicity from Step 1.

   Example

   \lambda=4 has algebraic multiplicity 2 but geometric multiplicity only 1: \underbrace{2}_{\text{algebraic}} \;>\; \underbrace{1}_{\text{geometric}} So A is defective — the search for P stops here.

   • geometric multiplicity equals algebraic multiplicity — this eigenvalue supplies enough independent eigenvectors; if every eigenvalue passes, continue to Step 4.
   • geometric multiplicity is strictly less than algebraic multiplicity — A is defective; stop here, no invertible P exists.

   Why: an eigenvalue can never supply MORE independent eigenvectors than its algebraic multiplicity, but it can supply fewer, and one shortfall anywhere sinks diagonalizability for the whole matrix, even if every other eigenvalue behaves perfectly.

4. 4

   Add up the geometric multiplicities across every eigenvalue (counting 1 automatically for each distinct one); A is diagonalizable exactly when that total is n.

   Example

   A matrix with two distinct eigenvalues (like the one in the next worked example) automatically totals 1+1=2=n independent eigenvectors — diagonalizable with no eigenspace check ever needed. A above never reaches this step: it already stopped at Step 3, defective.

   Why: this is the exact count the diagonalize-a-matrix recipe's own independence check needs — this test just carries it out carefully for a repeated eigenvalue instead of assuming it.