Proving that the orthogonal projection matrix is symmetric, and has eigenvalues of $0$ and $1$.

Question

We have been asked to prove the following things about orthogonal projection matrices:

Let $V \subseteq \mathbb R^d, \dim(V)=k$ and let $v_1, v_2,\dots,v_k$ be an orthonormal basis of $V$. Let $P=\sum_{i=1}^kv_iv_i^\top$ (Notice this is an outer product). Show that :

1. $P$ is symmetric.
2. The eigenvalues of $P$ are $0$ or $1$ and that $v_1,\dots, v_k$ are the eigenvectors corresponding to the eigenvalue $1$.

I think I got the symmetric part figured out, but the second subsection is giving me a really hard time... Anyone with a hint or a direction?

Answer 1

Hint: For part 1, recall that for matrices $A,B$, we have $(AB)^T = B^T A^T$ whenever $AB$ is defined and $(A + B)^T = A^T + B^T$ whenever $A+B$ is defined. Alternatively, start by convincing yourself that the matrix $vv^T$ is symmetric for any column vector $v$.

For part 2, notice that $(vv^T)x = v(v^Tx) = (v^Tx) \cdot v$. With that in mind, show that if $x$ is an element of $V$, then $x$ is an eigenvector of $P$. (What is the corresponding eigenvalue?) Similarly, if $x$ is an element of $V^\perp$ (the orthogonal complement to $V$), then $x$ is an eigenvector of $P$. (What is the corresponding eigenvalue?)

Now, argue that we can select a complete set of eigenvectors (that is, eigenvectors that form a basis of $\Bbb R^d$) consisting of vectors either from $V$ or $V^\perp$.

Answer 2

It follows from projection definition that the polynom $P(X) = X(1-X)$ verify $P(p_V) = 0$ . Since it has distinct two simple roots, $p_V$ is diagonalizable and has eigenvalues in $\{0, 1\}$.

For symmetry : Use $E = V \bigoplus V^{\perp}$, we have :

$(p_V(x) , y ) = (p_V(x) , p_V(y)+p_{V^\perp}(y) ) = (p_V(x) , p_V(y) ) = (p_V(x)+p_{V^\perp}(x), p_V(y) ) =(x, p_V(y) ) $