How to prove that if $A$ is a square matrix on $\mathbb{R}$, $A$ is nilpotent, then trace($A$)=0

Question

I need some help, I need to prove the following: Let $A$ be a square matrix on $\mathbb{R}$, if $A$ is nilpotent then Trace($A$)=0

I have seen some results for complex entries of the matrix, but what happens with real entries?

I write this because it has been difficult to understand the other answers some people have written on this website, thanks!

Answer 1

Observe that $A\in M(n,\mathbb R)\subset M(n,\mathbb C)$ So trace of $A$ remains the same whether we view it as an element of $M(n,\mathbb R)$ or of $M(n,\mathbb C)$. Lets us see $A$ as a matrix from $M(n,\mathbb C)$.Then Trace of a matrix is sum of all the eigenvalues of the matrix taken with multiplicity . (all the n eigenvalues of A)

Eigenvalues of nilpotent matrices are zero.(why?)

$A$ is a nilpotent matrix that is $\exists m>0$ such that $A^m=0$. If $\lambda$ is an eigenvalue of $A$ then there is a vector $v \not = 0$ such that $$A(v)=\lambda v \implies A^m(v)=\lambda^mv=0 \implies \lambda=0$$ So all Eigenvalues are zero.