Vtyjkyuk

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 $(lambda-1)^mlambda^m-n$




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 $BbbR^n=E_0 oplus E_1$ 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 $mne n$?

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.

$A$ is diagonalizable iff $textdim(E_lambda i)=textmultiplicity of $lambda_i$ $, for all $i$

By hypothesis, dimension of $N(A-I)=m$ means $textdim(E_1)=m$=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 $$rho_A(x)= (lambda-1)^m (lambda)^n-m$$

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

Hence $D = operatornamediag(underbrace1, ldots, 1_m, underbrace0, ldots, 0_n-m)$ or a permutation thereof, so the characteristic polynomial of $A$ is $$det(A - lambda I) = det(D - lambda I) = det operatornamediag(underbrace1-lambda, ldots, 1-lambda_m, underbrace-lambda, ldots, -lambda_n-m) = (-1)^n (lambda-1)^mlambda^n-m$$