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?
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?
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.