Exercise 2128Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

Trigonalization and eigenvalues

#trigonalization#characteristic polynomial#real eigenvalues

Statement

Let $A$ be an $n\times n$ real matrix. Show that $A$ is trigonalizable over $\mathbb{R}$ if and only if its characteristic polynomial has all roots in $\mathbb{R}$ .

Hint

Use induction on $n$ : find a real eigenvalue, which gives a stable line, then apply the induction hypothesis to the quotient.

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Solution

Forward direction. If $A$ is similar to an upper triangular matrix $T$ , then $\chi_A=\chi_T$ , which is a product of linear factors, so all roots are real.

Converse. Suppose $\chi_A$ has all roots in $\mathbb{R}$ . Then $A$ has at least one real eigenvalue $\lambda_1$ , hence a stable line $L_1=\textrm{span}(v_1)$ with $Av_1=\lambda_1 v_1$ . Complete $(v_1)$ to a basis of $\mathbb{R}^n$ . In this basis, the matrix of $A$ is block upper triangular:
$M = \begin{pmatrix} \lambda_1 & * \\ 0 & B \end{pmatrix}.$
The characteristic polynomial of $B$ divides $\chi_A$ , so it also has all roots in $\mathbb{R}$ . By induction, $B$ is trigonalizable, and consequently so is $M$ .

Conclusion. Trigonalizability over $\mathbb{R}$ is equivalent to all eigenvalues being real.

Trigonalization and eigenvalues

Statement

Solution

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