Exercise 1665Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

A matrix with a single eigenvalue

#eigenvalues#eigenspace#diagonalization

Statement

Let
$A = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}.$
Find the eigenvalues of $A$ and $\dim E_3$ . Is $A$ diagonalizable?

Hint

The matrix is upper triangular with diagonal $3,3,3$ . Compute $\ker(A-3I)$ .

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Eigenvalue. $A$ is upper triangular with all diagonal entries equal to $3$ , so the only eigenvalue is $\lambda=3$ , algebraic multiplicity $3$ .

Eigenspace.
$A-3I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.$
The kernel is $\{(x,0,0)\}$ , dimension $1$ .

Conclusion. Geometric multiplicity $1<3$ (algebraic multiplicity), so $A$ is not diagonalizable.

Remark. $A$ is a single Jordan block of size $3$ for $\lambda=3$ .