Let $V$ be a finite dimensional space and let $T, N\in L(V)$ such that $N$ is nilpotent and $NT=TN$. Show that
(a) $T$ is invertible iff $T+N$ is invertible
(b) $\det(T)=det(T+N)$ and $p_T(t)=p_{T+N}(t)$
To show all of this I'm trying to do the following:
I think it's enought to prove that $p_T(t)=p_{T+N}(t)$ because if $p_T(t)=p_{T+N}(t)$ then we have that, in particular, the constant terms of the polynomials are the same but the constant term of $p_T(t)$ is $(-1)^n\det(T)$ and of $p_{T+N}(t)$ is $(-1)^n\det(T+N)$, then $(-1)^n\det(T)=(-1)^n\det(T+N)$ and so $\det(T)=\det(T+N)$.
Now, as we know $\det(T)=\det(T+N)$, then "$T$ is invertible iff $T+N$ is invertible".
So, I think it's enought to prove $p_T(t)=p_{T+N}(t)$. Is this correct?
Attempt to show $p_T(t)=p_{T+N}(t)$:
I'm trying to show $T$ and $T+N$ have the same eigenvalues. We know the eigenvalues of a nilpotent polynomial are only zero.
let $\lambda$ be a eigenvalue of $T+N$ with eigenvector $v$, then, we have $\lambda v=(T+N)(v)=T(v)+N(v)$ we need to have then that $T(v)=\lambda_1 v$ and $N(v)=\lambda_2 v$ but $\lambda_2=0$ because $N$ is nilpotent, then every eigenvalue of $T+N$ is also a eigen value of $T$.
On the other hand, if $\lambda v=T(v)$, then $\lambda v=(T+N)(v)=T(v)+N(v)=\lambda v+0\cdot v$.
Is everything correct (I'm not using that $T$ and $N$ commute)? Thanks!
You are correct in thinking that proving $p_T(t) = p_{T+N}(t)$ is enough, and that proving $T$ and $T+N$ have the same eigenvalues will suffice. However, your proof is wrong, and as you mentioned you never used the fact that $T$ and $N$ commute, so you know your proof can't be right.
The mistake in your proof comes from thinking that $\lambda_2 = 0$. This would be the case if $v$ was an eigenvector of $N$, but it is actually an eigenvector of $T+N$, so as far as you know $Nv$ doesn't have to be $0$.
As for how to actually prove it, a sketch of how I would do it is by using the fact that (1) commuting matrices are simultaneously upper triangularizable, (2) adding upper triangular matrices adds their eigenvalues, and (3) similar matrices have the same eigenvalues. I'm not sure how else to do it if you haven't learned fact (1) yet, other than essentially proving fact (1) yourself. Hope this was helpful.