Exercise 1771Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026

The minimal polynomial of a matrix

#minimal polynomial#characteristic polynomial#diagonalizability

Statement

Compute the minimal polynomial of
$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}.$

Hint

The characteristic polynomial is $(X-2)^2$ . The minimal polynomial divides it, so it is either $X-2$ or $(X-2)^2$ .

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Characteristic polynomial. Since $A$ is upper triangular with diagonal entries $2$ , $\chi_A(X) = (X-2)^2$ .

Test $X-2$ . Compute $A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \neq 0$ . So $X-2$ is not the minimal polynomial.

Test $(X-2)^2$ . $(A-2I)^2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}^2 = 0$ . So $(X-2)^2$ annihilates $A$ .

Conclusion. The minimal polynomial is $\mu_A(X) = (X-2)^2$ . The matrix is not diagonalizable because the minimal polynomial has a multiple root.