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.
If every eigenvalue of $A$ is zero, does this mean $A$ is a zero matrix?
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linear-algebra
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2I 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. – 2012-03-28
2 Answers
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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.
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0@e$x$ample Oh, well that was an accident, but you're welcome! – 2012-03-28
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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:
This can be proved by finding eigenvalue decomposition of $P^k$