How does the characteristics polynomial be $(\lambda-1)^m \lambda^{m-n}$ where $m \le n$ here?

Question

Question. Let $A$ be a 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:

1. The characteristic polynomial of $A$ is $(\lambda-1)^m\lambda^{m-n}$

2. $A^k=A^{k+1}$ for all positive integers $k$.

3. The rank of $A$ is $m$.

My Solution.

Since $A$ is symmetric matrix so there exists some non singular matrix, $P$ and a diagonal matrices $D$ whose diagonal entries are only $0$ and $1$ (as that of eigen values of $A$) so that $A=PDP^{-1}$. Therefore $A^2=PD^2P^{-1}=PDP^{-1}=A$ This implies $2$ is true.

Now according to the given condition $\dim E_1=m$.(here, $E_\lambda=$ eigenspace corresponding to $\lambda$). Now since $A$ is diagonalizable so $\Bbb{R}^n=E_0 \oplus E_1$ this implies $\dim E_0=n-m$. So $\dim Ker A=n-m$ i.e. $r(A)=m$. This prove $3$ to be true

But I cannot see any way to conclude 1.

Infact, I have a confusion about option $1$. here how does the power of $\lambda$ is $m-n$ in option 1? I mean $m \le n$ here. So how does $(\lambda-1)^m \lambda^{m-n}$ is a polynomial when $m\ne n$?

Please hep me to clear the confusion and option 1. Also let me know whether I made any mistake in all other options. Thank you ..

EDIT: Here the option 1 has been proved true (assuming the option 1 as it is given here). But I really don't get that answer.

Answer 1

$A$ is diagonalizable iff $\text{dim}(E_{\lambda i})=\text{multiplicity of $\lambda_i$ }$, for all $i$

By hypothesis, dimension of $N(A-I)=m$ means $\text{dim}(E_{1})=m$=multiplicity of $1$

So $(\lambda-1)^m$ is a factor of $\rho_A(x)$ and so $0$ is the eigenvalue of multiplicity $n-m$.

Thus $$\rho_A(x)= (\lambda-1)^m (\lambda)^{n-m}$$

Answer 2

$A$ diagonalizes to a diagonal matrix $D$ with only $0$ and $1$ on the diagonal. We know that $$n-m = \operatorname{rank}(A - I) = \operatorname{rank}(D - I) = \text{ number of zeroes on the diagonal of }D$$

Hence $D = \operatorname{diag}(\underbrace{1, \ldots, 1}_m, \underbrace{0, \ldots, 0}_{n-m})$ or a permutation thereof, so the characteristic polynomial of $A$ is $$\det(A - \lambda I) = \det(D - \lambda I) = \det \operatorname{diag}(\underbrace{1-\lambda, \ldots, 1-\lambda}_m, \underbrace{-\lambda, \ldots, -\lambda}_{n-m}) = (-1)^n (\lambda-1)^m\lambda^{n-m}$$