Exercise 1660Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

An involution is diagonalizable

#involution#diagonalization#eigenvalues

Statement

Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be a linear map with $f^2=\mathrm{id}$ . Show that $f$ is diagonalizable and determine its eigenvalues.

Hint

$f^2-\mathrm{id}=0$ , so $(f-\mathrm{id})(f+\mathrm{id})=0$ . The minimal polynomial divides $X^2-1=(X-1)(X+1)$ .

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Solution

Minimal polynomial. $f^2=\mathrm{id}$ implies $f^2-\mathrm{id}=0$ , so the minimal polynomial $\mu_f$ divides $X^2-1$ .

Eigenvalues. The roots of $\mu_f$ are among $1$ and $-1$ . If $f=\mathrm{id}$ , the only eigenvalue is $1$ ; if $f=-\mathrm{id}$ , the only eigenvalue is $-1$ ; otherwise both are eigenvalues.

Diagonalizability. $X^2-1$ has simple roots (in characteristic $\neq2$ ), so $\mu_f$ also has simple roots and $f$ is diagonalizable.

Eigenspaces. $E_1=\ker(f-\mathrm{id})$ , $E_{-1}=\ker(f+\mathrm{id})$ , and $\mathbb{R}^n=E_1\oplus E_{-1}$ .

Conclusion. Any involution is diagonalizable with eigenvalues in $\{1,-1\}$ .

An involution is diagonalizable

Statement

Solution

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