Let $A$ be real symmetric $n\times n$ matrix whose only eigenvalues are $0$ and $1$. Let the dimension of the null space of $A-I$ be $m$. Pick out the true statements.
The characteristic polynomial of $A$ is $(\lambda-1)^m(\lambda)^{m-n}$.
$A^k = A^{k+1}$
The rank of $A$ is $m$.
This is what I did: I found geometric multiplicity corresponding to eigenvalue value $0$ to be $n-m$(using symmetric matrix is diagonalizable ) while geometric multiplicity of eigenvalue value 1 is $m$(that is given) .so the characteristic polynomial of $A$ should be $(\lambda-1)^m(\lambda)^{n-m}$. While I am not sure about other two statements. Any kind of help is highly appreciated. Thanks.
Best Answer
As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$. Indeed, $\dim\ker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $\ker(A-I)=\ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.