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a) Characteristic polynomial will be $(\lambda-1)^m\lambda^{n-m}=0$?

b) as minimal polynomial of $\lambda(\lambda-1)=0$ so $2$ is true

c) $\dim Ker(A-I)=m$ so rank of $A$ is $m$

Am I correct?

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Does (a) in the question have a typo? The polynomial they give doesn't have degree $n$. – wj32 Oct 27 '12 at 10:16
I am sure for that, It is a question from a exam paper, I am sure that is typo – Un Chien Andalou Oct 27 '12 at 10:17
up vote 2 down vote accepted

A symmetric matrix whose only eigenvalues are $0$ and $1$ is a projection matrix, so $A^2=A$. Therefore (b) is correct.

Moreover, the range of $A-I$ coincides with the null space of $A$ (because $A$ is a projection) and this shows that the dimension of the range of $A$ is $m$, so the rank of $A$ is $m$, and this shows that (a) and (c) are also true.

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