# eigenvalues of a matrix with zero $k^{th}$ power

For a matrix $A$, where $A^k=0$, $k\ge1$, need prove that $trace(A)=0$; i.e sum of eigenvalues is zero. How do you approach this problem?

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Take an arbitrary eigenvector and apply $A^k$ to it. What does that tell you about the eigenvalues of $A$? –  Mark Bennet Feb 10 '13 at 21:13

hint: $A$ is a real matrix but you assume $A$ is a complex matrix and $f(x)=(x-a_1)(x-a_2)...(x-a_n)$ is its characteristic polynomial in the complex field. By induction you can prove that trac($A^k$)=$\sum_{i=1}^n {a_i^k}$ , and if $\forall k\in\mathbb N$ trac($A^k$)=$\sum_{i=1}^n {a_i^k}=0$ then $a_i=0$. Hence, $f(x)=x^n$, $\forall i$ $a_i=0$ $$trac A=\sum_{i=1}^n {a_i}=0$$

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I assume your matrix is an $n\times n$ matrix with, say, complex coefficients.

Since $A^k=0$, the spectrum of $A$ is $\{0\}$ (or the characteristic polynomial of $A$ is $X^n$). Next we can find an invertible matrix $P$ such that $PAP^{-1}$ is upper-triangular with $0$'s on the diagonal. So $$\mbox{trace}A=\mbox{trace}(PAP^{-1})=0$$ where we use the fact that $\mbox{trace} (AB)=\mbox{trace}(BA)$ in general.

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Thank you all. I got that. I appreciate your responses. –  user25004 Feb 10 '13 at 21:57

Do you know that the trace of a matrix is the sum of it's eigenvalues counted with multiplicity? Think about the Jordan Canonical Form of your matrix to see why this is so.

Now for a matrix that satisfies $A^k = 0$ prove that $0$ is its only eigenvalue.

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