25
Nov 18

## Eigenvalues and eigenvectors of a projector

### Eigenvalues and eigenvectors of a projector

Exercise 1. Find eigenvalues and eigenvectors of a projector.

Solution. We know that a projector doesn't change elements from its image: $Px=x$ for all $x\in\text{Img}(P).$ This means that $\lambda =1$ is an eigenvalue of $P.$ Moreover, if $\{x_i:i=1,...,\dim\text{Img}(P)\}$ is any orthonormal system in $\text{Img}(P),$ each of $x_i$ is an eigenvector of $P$ corresponding to the eigenvalue $\lambda =1.$

Since $P$ maps to zero all elements from the null space $N(P),$ $\lambda =0$ is another eigenvalue. If $\{y_i:i=1,...,\dim N(P)\}$ is any orthonormal system in $N(P),$ each of $y_i$ is an eigenvector of $P$ corresponding to the eigenvalue $\lambda =0.$

A projector cannot have eigenvalues other than $0$ and $1.$ This is proved as follows. Suppose $Px=\lambda x$ with some nonzero $x.$ Applying $P$ to both sides of this equation, we get $Px=P^2x=\lambda Px=\lambda ^2x.$ It follows that $\lambda x=\lambda^2x$ and (because $x\neq 0$) $\lambda =\lambda^2.$ The last equation has only two roots: $0$ and $1.$

We have $\dim\text{Img}(P)+\dim N(P)=n$ because $R^n$ is an orthogonal sum of $N(P)$ and $\text{Img}(P)$.  Combining the systems $\{x_i\},$ $\{y_i\}$ we get an orthonormal basis in $R^{n}$ consisting of eigenvectors of $P$.

### Trace of a projector

Recall that for a square matrix, its trace is defined as the sum of its diagonal elements.

Exercise 2. Prove that $tr(AB)=tr(BA)$ if both products $AB$ and $BA$ are square. It is convenient to call this property trace-commuting (we know that in general matrices do not commute).

Proof. Assume that $A$ is of size $n\times m$ and $B$ is of size $m\times n.$ For both products we need only to find the diagonal elements:

$AB=\left(\begin{array}{ccc} a_{11}&...&a_{1m}\\...&...&...\\a_{n1}&...&a_{nm}\end{array} \right)\left(\begin{array}{ccc} b_{11}&...&b_{1n}\\...&...&...\\b_{m1}&...&b_{mn}\end{array} \right)=\left(\begin{array}{ccc} \sum_ia_{1i}b_{i1}&...&...\\...&...&...\\...&...&\sum_ia_{ni}b_{in}\end{array} \right)$

$BA=\left(\begin{array}{ccc} b_{11}&...&b_{1n}\\...&...&...\\b_{m1}&...&b_{mn}\end{array} \right)\left(\begin{array}{ccc} a_{11}&...&a_{1m}\\...&...&...\\a_{n1}&...&a_{nm}\end{array} \right)=\left(\begin{array}{ccc} \sum_ja_{j1}b_{1j}&...&...\\...&...&...\\...&...&\sum_ja_{jm}b_{mj} \end{array}\right)$

All we have to do is change the order of summation:

$tr(AB)=\sum_j\sum_ia_{ji}b_{ij}=\sum_i\sum_ja_{ji}b_{ij}=tr(BA).$

Exercise 3. Find the trace of a projector.

Solution. In Exercise 1 we established that the projector $P$ has $p=\dim\text{Img}(P)$ eigenvalues $\lambda =1$ and $n-p$ eigenvalues $\lambda =0.$ $P$ is symmetric, so in its diagonal representation $P=UDU^{-1}$ there are $p$ unities and $n-p$ zeros on the diagonal of the diagonal matrix $D.$ By Exercise 2

$tr(P)=tr(UDU^{-1})=tr(DU^{-1}U)=tr(D)=p$.