Exercise 1661Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

A projection is diagonalizable

#projection#diagonalization#eigenvalues

Statement

Let $p:\mathbb{R}^n\to\mathbb{R}^n$ be a projection, i.e. $p^2=p$ . Show that $p$ is diagonalizable and determine its eigenvalues.

Hint

If $p^2=p$ , then $p(p-\mathrm{id})=0$ , so the minimal polynomial of $p$ divides $X(X-1)$ .

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Minimal polynomial. From $p^2=p$ : $p^2-p=0$ , so $p(p-\mathrm{id})=0$ . The minimal polynomial $\mu_p$ divides $X(X-1)$ .

Eigenvalues. The possible eigenvalues are roots of $\mu_p$ : $0$ and $1$ .

Diagonalizability. The minimal polynomial has simple roots (no repeated factor), so $p$ is diagonalizable. Its eigenvalues are $0$ and $1$ .

Eigenspaces. $E_1=\textrm{Im} p$ (fixed points), $E_0=\ker p$ . Moreover $\mathbb{R}^n=E_0\oplus E_1$ .

Conclusion. Any projection is diagonalizable with eigenvalues $0$ and $1$ .