Exercise 933Linear algebra › Eigenvalues and diagonalisationHigh schoolAdded 3 Nov 2026

A 2×2 matrix with a double eigenvalue

#eigenvalues#double root#diagonalizability

Statement

Find the eigenvalues and eigenvectors of
$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}.$
Is $A$ diagonalizable?

Hint

The characteristic polynomial has a double root at $2$ . Compute the eigenspace $\ker(A-2I)$ .

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Characteristic polynomial.
$\det(A-\lambda I) = (2-\lambda)^2.$
So the only eigenvalue is $\lambda=2$ , with algebraic multiplicity $2$ .

Eigenspace.
$A-2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.$
The kernel is given by $y=0$ , so $E_2=\textrm{span}((1,0))$ , dimension $1$ .

Conclusion. The geometric multiplicity of $\lambda=2$ is $1<2$ (the algebraic multiplicity), so $A$ is not diagonalizable.

Remark. The matrix is a single Jordan block of size $2$ for the eigenvalue $2$ .