25
Nov 18

Eigenvalues and eigenvectors of a projector

Eigenvalues and eigenvectors of a projector

Exercise 1. Find eigenvalues 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.

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