T ∈ L(V) is nilpotent if and only if σ(T) = {0}.

Question

I know that T is nilpotent if and only if only eigenvalues of T is 0.

How can we prove the statement $T ∈ L(V)$ is nilpotent if and only if $\sigma (T) = \{0\}$ where $\sigma (T)$ denotes singular value of T?

Answer 1

This is not true, $A =\pmatrix{ 0 & 1\\0 & 0 }$ is nilpotent, but has non-zero singular value $\sigma_1=1$.

Answer 2

I think that by "singular value" you mean "eigenvalue".

1. Let $T$ be nilpotent. Then $T^m=0$ for some $m \in \mathbb N.$ If $ \lambda$ is an eigenvalue of $T$ then $ \lambda^m$ is an eigenvalue of $T^m=0.$ This gives $ \lambda=0.$

2. If $ \sigma(T)=\{0\},$ then the char. polynomial of $T$ is given by $p(x)=x^n,$ where $ n= \dim V.$ Cayley - Hamilton now says that $0=p(T)=T^n.$