Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite
1

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.

share|improve this question
2  
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
2  
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

1 Answer

up vote 15 down vote accepted

No, any strictly upper triangular matrix, such as:

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

will have all eigenvalues zero.

share|improve this answer
1  
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
1  
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
1  
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
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.