How can I prove that the linear transformation has at most one non-zero eigen value?

Question

My Attempt: If $0$ was not an eigenvalue of $T$ then the eigenvalues of $T^k$ would have been non-zero also, $\ker(T^n) = \ker(T^{n-1}) = \mathbf{0}$.

Am I correct?

Can anyone please help me with (b)?

Answer 1

(a) : Because $\mathrm{Ker}(T^{n-2}) \subset \mathrm{Ker}(T^{n-1})$, the hypothesis can be rewritten as : $\exists x \in \mathrm{Ker}(T^{n-1}) \setminus \mathrm{Ker}(T^{n-2})$. For such a $x$, you have $T^{n-2}x \neq 0$, and $T(T^{n-2}x)=0$, so $0$ is an eigenvalue of $T$ (and $T^{n-2}x$ an eigenvector associated to the eigenvalue $0$).

(b) : For this part, I would use the classical result on iterated kernels. If $\mathrm{Ker}(T^{n-2}) \varsubsetneq \mathrm{Ker}(T^{n-1})$, that means that $$\lbrace 0 \rbrace \varsubsetneq \mathrm{Ker}(T) \varsubsetneq ... \varsubsetneq \mathrm{Ker}(T^{n-2}) \varsubsetneq \mathrm{Ker}(T^{n-1})$$

You deduce that $\mathrm{dim}(\mathrm{Ker}(T^{n-1})) \geq n-1$. You can see (by trigonalizing $T$ for example), that $T$ cannot have more than one non-zero eigenvalue.