# 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

No, any strictly upper triangular matrix, such as:

$$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

will have all eigenvalues zero.

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Jordan blocks with zero eigenvalues (1's in the superdiagonal and zeros everywhere else) are a classical example; in fact any strictly upper triangular matrix is similar to a Jordan block with zero eigenvalues, or direct sums of such Jordan blocks. –  Ｊ. Ｍ. Mar 28 '12 at 21:23
any strictly (diagonal = 0) triangular matrix to be precise. no matter whether upper or lower –  example Mar 28 '12 at 21:24
Indeed. But there are more examples than that even. –  Matt Pressland Mar 28 '12 at 21:25
I see. Since every eigenvalue of A is zero, then there exists an orthogonal matrix P such that B=(P^T)AP=(P^-1)AP is upper triangular, then B is similar to A. Which means B is upper triangular with zeros on the mail diagonal. Consequently, A is nilpotent. –  Shannon Mar 28 '12 at 21:42
@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