# If every eigenvalue of $A$ is zero, does this mean $A$ is a zero matrix?

If every eigenvalue of $A$ is zero, show that $A$ is nilpotent. I got this question as my homework. I am just wondering if every eigenvalue of $A$ is zero, then $A$ is zero, why bother to prove $A$ is nilpotent.

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For the statement you actually want to prove (every eigenvalue is $0$ implies the matrix is nilpotent) you need to be working over an algebraically closed field, e.g., $\mathbb{C}$. –  Arturo Magidin Mar 28 '12 at 21:22
This is likely a dupe, but I am unable to find it. –  Aryabhata Mar 28 '12 at 21:24
I think your confusion might come from the fact that if it were the case that all eigenvalues are 0, and your matrix $A$ is diagonalisable, then you would have $A=P^{-1}0P=0$. But in general, your matrix won't be diagonalisable. –  M Turgeon Mar 28 '12 at 21:37
@MTurgeon Thanks, I just realized this 2 min ago. –  Shannon Mar 28 '12 at 21:44

$$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
@Matt: you are right. But for the life of me I cannot find a general criterion other than $A^\infty=0$. If $A$ and $c A^T$ have no elements other then zeroes in common (for any $c$) then there cannot be any two-circles (ie eigenstates with two values differeing from 0), but there might very well be some with three or more... –  example Mar 28 '12 at 21:43