Orthogonal Projections and eigenvalues

Question

Let $V$ be a subspace of $\mathbb{R}^n$ with $V \ne \mathbb{R}^n$ and $V\ne \{0\}$. Let $A$ be the matrix of the linear transformation $\text{proj}_V : \mathbb{R}^n\to\mathbb{R}^n$ that is the projection onto $V$.

(a) Give two real numbers that are eigenvalues of $A$.

I think that one of the eigenvalues will be $0$ due to the fact that if $V$ is a subspace the orthogonal complement will also be a subspace. And projection takes any vector to $0$ so I believe $0$ will be one? But I don’t know if this is the correct logic and stuck on what the other will be.

Answer 1

You are right indeed

• for any $v\in V^\perp$ we have $Pv=0 \implies \lambda =0$

• for any $v\in V$ we have $Pv=v \implies \lambda =1$

Answer 2

Hint:

$A$ satisfies $A^2 = A$ or $A(A-I) = 0$ so the eigenvalues are contained in $\{0,1\}$.

For $x \in V, x\ne 0$ we have $Ax = x$, and for $x \in V^\perp, x\ne 0$ we have $Ax = 0$ so $0$ and $1$ are indeed eigenvalues.