Let $V$ be a finite-dimensional vector space and $T:V\rightarrow V$ be an operator such that $tr(\Lambda^q(T))=0$ for all $q$. Prove that $T$ is nilpotent.
I would like to understand the following proof:
Proof: Operator $T$ is nilpotent if the spectrum of $T$ is a set of zeros, i.e. $\{0\}$. Let us assume $T$ has values $\mu_i$, where $i$ assumes real positive numbers $i>0$. Hence, for $\Lambda^q(T)=0$, then the value $i$ in the finite dimension $q$ must be 0, i.e. $\mu_i=0$ for all $1\leq i_1, i_2, \dots \geq q$, implying that $i$ will also be zero.