0
$\begingroup$

Given an $n \times n$ matrix with rank $m$, we can know that the algebraic multiplicity of the eigenvalues of such matrix is: for eigenvalue$=1$ $a.m=m$; for eigenvalue$=0$ $a.m=n-m$.

However, is it correct to assume that this is equivalent to their geometric multiplicities?

$\endgroup$
1
$\begingroup$

Yes. You can see this directly with the rank-nullity theorem: if $P$ has rank $m$, then because $P^2=P$ you get that every vector in the image is an eigenvector with eigenvalue $1$. Here's the proof: if $b \in C(P)$, then $b=Px$, now $Pb=P^2x=Px=b$, so $b$ is an eigenvector with eigenvalue $1$.

So the eigenvalue $1$ has geometric multiplicity $m$. Then the rank-nullity theorem tells you the nullity is $n-m$, i.e. the geometric multiplicity of the eigenvalue $0$ is $n-m$.

Note that this actually does not depend on $P$ being an orthogonal projector, it only requires $P^2=P$.

$\endgroup$
  • $\begingroup$ ohh I see. Thank you very much! $\endgroup$ – tuso96 Jan 14 '15 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.