Exercise 1664Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

Geometric versus algebraic multiplicity

#multiplicities#eigenspace#diagonalization

Statement

Let $A$ be an $n\times n$ matrix. Show that for every eigenvalue $\lambda$ ,
$1\le\dim E_\lambda\le m_\lambda,$
where $m_\lambda$ is the algebraic multiplicity of $\lambda$ (as root of $\chi_A$ ).

Hint

The first inequality is because an eigenvalue has a nonzero eigenvector. For the second, use the rank-nullity theorem and properties of the characteristic polynomial.

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Solution

Lower bound. By definition, an eigenvalue has at least one nonzero eigenvector, so $\dim E_\lambda\ge1$ .

Upper bound. Let $k=\dim E_\lambda$ and choose a basis $(v_1,\ldots,v_k)$ of $E_\lambda$ . Extend this basis to a basis of $\mathbb{R}^n$ . In this basis, the matrix of $A$ is block upper triangular:
$\begin{pmatrix} \lambda I_k & * \\ 0 & * \end{pmatrix}.$
Therefore $\chi_A$ has the factor $(\lambda-X)^k$ , i.e. the algebraic multiplicity of $\lambda$ is at least $k$ . Hence $k\le m_\lambda$ .

Conclusion. $1\le\dim E_\lambda\le m_\lambda$ . Equality for all $\lambda$ iff $A$ is diagonalizable.

Geometric versus algebraic multiplicity

Statement

Solution

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