Exercise 1659Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

The diagonalizability criterion

#diagonalizability#multiplicity#eigenspaces

Statement

Let $A$ be an $n\times n$ matrix. Show that $A$ is diagonalizable if and only if:
1. the characteristic polynomial splits into linear factors over $\mathbb{R}$ ;
2. for each eigenvalue $\lambda$ , its geometric multiplicity $\dim E_\lambda$ equals its algebraic multiplicity (the multiplicity of $\lambda$ as a root of $\chi_A$ ).

Hint

If $A$ is diagonalizable, it is similar to a diagonal matrix $D$ , and the eigenspaces of $A$ correspond to those of $D$ via the change of basis.

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Solution

Forward direction. If $A=PDP^{-1}$ with $D$ diagonal, the diagonal entries of $D$ are the eigenvalues of $A$ , and each eigenvalue $\lambda$ appears on the diagonal exactly as many times as its algebraic multiplicity. The corresponding eigenspace has the same dimension as the eigenspace of $D$ for $\lambda$ , which is the number of times $\lambda$ appears on the diagonal. So geometric multiplicity equals algebraic multiplicity.

Converse. Suppose the characteristic polynomial splits and each geometric multiplicity equals algebraic multiplicity. Then the sum of the geometric multiplicities equals the sum of the algebraic multiplicities, which equals $n$ . Since the eigenspaces for distinct eigenvalues are in direct sum, the union of bases of the eigenspaces is a basis of $\mathbb{R}^n$ , so $A$ is diagonalizable.

Conclusion. The two conditions are necessary and sufficient.

The diagonalizability criterion

Statement

Solution

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