Recipe

Determine whether a matrix is diagonalizable, via multiplicities

Use when: you have a square matrix's eigenvalues and need to know whether it's diagonalizable, without necessarily building $P$ and $D$

Goal: a definite conclusion, diagonalizable or not, backed by an exact multiplicity comparison at every eigenvalue

Taught in Why diagonalization can fail

Quick recipe

Find each eigenvalue's algebraic multiplicity
Find each eigenvalue's geometric multiplicity
Compare the two multiplicities at every eigenvalue

1
Factor the characteristic polynomial and read off each root $\lambda$ 's exponent; that exponent is its algebraic multiplicity.

Example

For $A=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ : $\det(A-\lambda I)=(2-\lambda)^2$ , so $\lambda=2$ has algebraic multiplicity $2$ .

Why: algebraic multiplicity is defined directly as a root's exponent in the characteristic polynomial, so this step is just reading the factored polynomial.
2
Solve $(A-\lambda I)\mathbf v=\mathbf0$ and count the dimension of the resulting eigenspace; that dimension is the geometric multiplicity.

Example

$(A-2I)\mathbf v=0$ gives $\begin{pmatrix}0&1\\0&0\end{pmatrix}\mathbf v=0$ , so $v_2=0$ ; the eigenspace is $\text{span}\{(1,0)\}$ , dimension $1$ .

⚡ Shortcut

Whenever an eigenvalue's algebraic multiplicity is $1$ , its geometric multiplicity is automatically $1$ too, since $1\le\text{geometric}\le\text{algebraic}$ pins it down with no computation. For a $3\times3$ matrix's eigenvalue $\lambda=5$ with algebraic multiplicity $1$ , the geometric multiplicity is $1$ as well, with no system to solve.

Why: geometric multiplicity is defined as the eigenspace's dimension, and directly counting a basis for the null space of $A-\lambda I$ is how that dimension is actually found.
3
Check whether geometric equals algebraic for each eigenvalue in turn, not just the first one you compute.

Example

For a $3\times3$ matrix $C$ with $\lambda=2$ (algebraic $2$ , geometric $1$ ) and $\lambda=5$ (both $1$ , by the shortcut above): the gap at $\lambda=2$ alone already rules out diagonalizability.
- every eigenvalue has geometric equal to algebraic — $A$ is diagonalizable.
- even one eigenvalue has geometric strictly less than algebraic — $A$ is NOT diagonalizable, and the rest don't need checking.
Why: diagonalizability requires an independent eigenvector for every unit of algebraic multiplicity across the whole space, so a single gap anywhere breaks it entirely.

Common mistakes

Stopping after one eigenvalue's multiplicities match, without checking every other eigenvalue too.
Computing geometric multiplicity from the number of columns of $A-\lambda I$ instead of its null space's actual dimension.
Assuming any repeated eigenvalue forces a gap, when equality (as in $B=2I$ ) is genuinely possible.
Applying the algebraic-multiplicity- $1$ shortcut to an eigenvalue whose algebraic multiplicity is actually $2$ or more.