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?


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 $P^2=P$.

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

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