If $A$ is nilpotent matrix then $tr(A)=0$

Question

I want to prove if $A$ is nilpotent matrix then $tr(A)=0$. I have seen some answers on MSE about this question but none of them were clear enough to be understood for me. I appreciate if someone explains the proof in mathematical notation rather than a general explanation about it.

Answer 1

In general the trace of a matrix is the sum of the eigenvalues in the algebraic closure. Suppose $\lambda$ is an eigenvalue of $A$ and let $v\neq 0$ be such that $$Av=\lambda v$$ Suppose $A^n=0$ for some $n$. Then $$A^nv=\lambda^nv=0$$ Since $v\neq 0$, we have that $\lambda^n=0$, hence $\lambda=0$. Thus all eigenvalues of $A$ are zero hence the trace is zero.

Answer 2

Let $\lambda$ be an eigenvalue of $A$, and choose some associated eigenvector $v$. Now, we show $A^kv = \lambda^kv$. True for $k=1$, and then $A^nv = A^{n-1}(Av) = \lambda A^{n-1}v = \lambda^nv$ by induction. Now, assume our nilpotent matrix has some eigenvalue $\lambda \ne 0$. We know that $A^k = 0$ for some $k$, but then $\lambda^k \ne 0$ is an eigenvalue of the zero matrix, which is nonsense. Thus, all eigenvalues of $A$ are $0$, as well as their sum (the trace).

Answer 3

If $A\in\mathbb{K}^{n\times n}$ is nilpotent, then its characteristic polynomial is $\chi_A (\lambda)=(-1)^n\lambda^n$ and in general por any $M\in\mathbb{K}^{n\times n}$, $\chi_M (\lambda)=(-1)^n\lambda^n+(-1)^{n-1}(\text{trace }M)\lambda^{n-1}+\cdots+\det M$ so, $\text{trace }A=0$.

Answer 4

I will try to give some explanation of the proofs given earlier. First of all note that trace is cyclic which means $tr(ABC)=tr(CAB)=tr(BCA)$, which gives us following equality $tr(P^{-1}AP)=tr(A)$. So trace is invariant under change of basis.
Now as $A$ is nilpotent, there exist some positive integer $k$ such that $A^k=0$. Now because $A^k=0$, the kernel of $A^k$ is the whole space. Also note that if $v \in \ker(A^j)$ then $v\in \ker (A^{j+1})$, as $A^{j+1}v=AA^jv=0$. So we have the following relation $\{0\}=\ker A^0\subseteq\ker A^1\subseteq\ker A^2\subseteq\cdots \subseteq \ker A^k=V $. Choosing a basis of $\ker A^1$, then extending to a basis of the next space $\ker A^2$, and so on, eventually gives a basis of the whole space, because $\ker A^k=V$. By construction change of basis of $A$ to this new basis leads to a strictly upper triangular matrix, to see this note that for $v\in \ker A^j$, we have $A^jv=0 \rightarrow A^{j-1}Av=0 \rightarrow Av \in \ker A^{j-1}$. So for any $v \in \cup_{i=1}^j \ker A^i$, we have $Av \in \cup_{i=1}^{j-1} \ker A^i$. And so A is strictly upper triangular matrix in our constructed basis. So in this new basis we have trace equal 0 and because trace is invariant to change of basis, $tr(A)=0$.