Skeptical about an Elementry point concerning polynomials of linear operators

Question

Consider the following scenario. We have a linear operator $p(T) = c_0I+c_1T+c_2T^2+\cdot\cdot\cdot+c_mT^m$ over the finite dimensional vector space $V$ such that $p(T)v = 0$ for some non-zero $v\in V$.

Now lets say that we are able to factorize the aforementioned polynomial to yield $p(T) = \beta(T-\lambda_1I)(T-\lambda_2I)\cdot\cdot\cdot(T-\lambda_mI)$ where $\beta\neq 0$ consequently $\beta(T-\lambda_1I)(T-\lambda_2I)\cdot\cdot\cdot(T-\lambda_mI)v = 0$ does this then imply that $(T- \lambda_jI)v = 0$ for at least one $j\in\{1,2,....,n\}$ and if not why not?

I know this is an elementrary point but i seem to be having some trouble understanding it?

Answer 1

The answer is no if you insist on using the same $v$, as @mechanodroid has told us.

But it does imply that for at least one $j$ we have $(T-\lambda_jI)w = 0$ for some $w \in V$ with $w \ne 0$.

Indeed, if $(T-\lambda_2I)\cdots(T-\lambda_mI)v \ne 0$, we can take $j=1$ and $w=(T-\lambda_2I)\cdots(T-\lambda_mI)v$. Otherwise, we repeat the process with $(T-\lambda_2I)\cdots(T-\lambda_mI)v =0$.

Answer 2

It does not hold.

Let $T = \pmatrix{0 & 1 \\ 0 & 0}$.

We have that $T^2 = 0$ so in particular $T^2 e_2 = 0$. However, $Te_2 = e_1 \ne 0$.