Let $A$ be real symmetric $n\times n$ matrix whose only eigenvalues are 0 and 1. Pick out the true statements.

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.

• The first is true if we switch $m$ and $n$ in the power of $\lambda$, namely $p_A(\lambda)=(\lambda-1)^m\lambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.
• The second is false if $k=0$ unless $A=I$, but true for $k\geq 1$.
• The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.

To answer 2 and 3, consider diagonalizing $A$. And note that it is enough to show 2 for $k=1$.