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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.

Davide Giraudo
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Shannon
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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
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    This is likely a dupe, but I am unable to find it. – Aryabhata Mar 28 '12 at 21:24
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    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

2 Answers2

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No, any strictly upper triangular matrix, such as:

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

will have all eigenvalues zero.

mdp
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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. ain't a mathematician Mar 28 '12 at 21:23
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    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. – mdp Mar 28 '12 at 21:25
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    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
  • @example I'm afraid I'm not sure I can be much help there. My reasoning was merely that there must be matrices which are similar to strictly upper/lower triangular matrices without actually being triangular themselves. – mdp Mar 28 '12 at 21:51
  • @Matt (this comment was about an example that is not like a triangular matrix before I realized that...) any nilpotent matrix can be rewritten as a strictly triangular matrix by a change in basis (ie. exchange row 2 and 3 and column 2 and 3 to bring an element "into" the triangular area). Thanks for the impulse to this train of thought :D – example Mar 28 '12 at 21:59
  • @example Oh, well that was an accident, but you're welcome! – mdp Mar 28 '12 at 22:01
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MDP wrote that triangular matrix has all zero eigenvalues. The thing is that this is just a corollary from a more general statement.

Any nilpotent matrix has all zero eigenvalues

Nilpotent is the matrix which for some $k$ has $P^k = 0$.

For example for this non-obvious matrix is: enter image description here

This can be proved by finding eigenvalue decomposition of $P^k$

Salvador Dali
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