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Prove that a matrix with only zero eigenvalues must be nilpotent.

How will I be able to prove this?

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Also, note that the converse is true. – Tom Oldfield Dec 11 '12 at 4:09
You need to make assumptions on the base field for this to work. It won't work over $\Bbb R$ for instance. – Olivier Bégassat Dec 11 '12 at 4:12
up vote 4 down vote accepted

We have to assume that we are considering complex matrices. Over the reals the assertion is not true, as the example $$ \begin{bmatrix}0&0&0\\0&0&1\\0&-1&0 \end{bmatrix} $$ shows.

Over $\mathbb C$, one can do the Schur decomposition, where $A=VTV^*$, with $V$ unitary and $T$ upper triangular. Since the diagonal of $T$ has to contain the eigenvalues of $A$, it has be zero. And it is an easy exercise that if $T$ is an $n\times n$ upper triangular with diagonal zero, then $T^n=0$. So $A^n=(VTV^*)^n=VT^nV^*=0$. So $A$ is nilpotent.

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Thank you very much! I wish there was a way to rate the top two answers because you and Tom helped a lot on this question. – diimension Dec 11 '12 at 5:00
I'm confused, this matrix has non-zero eigenvalues, admittedly they're not in $\mathbb{R}$, but I wouldn't call it a counter-example. Certainly we may need to consider $A$ as a complex matrix to decompose it, and hence prove the theorem, but the result still holds for real $A$, doesn't it? – Tom Oldfield Dec 11 '12 at 5:20
If your field is $\mathbb R$, then the matrix above has only $0$ as an eigenvalue. If your field is $\mathbb C$, then the eigenvalues are $0$, $i$, $-i$. If your field is $\mathbb Z_2$, then the eigenvalues are $0,1$. The point of the example is that if your field is $\mathbb R$ then the assertion in the question is not true. It is the same kind of distinction where $x^2+1$ is irreducible over $\mathbb R$ and reducible over $\mathbb C$. – Martin Argerami Dec 11 '12 at 5:25

Hint: Look at the characteristic polynomial, then use Cayley-Hamilton.

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I have not learned Cayley-Hamilton theorem yet. – diimension Dec 11 '12 at 4:12

Cayley-Hamilton theorem says a linear transformation (equivalently, of course, its matrix) satisfies its own characteristic polynomial. What is the characteristic polynomial of a matrix with only zero eigenvalues?

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I have not learned Cayley-Hamilton theorem yet. – diimension Dec 11 '12 at 4:11

Hint: $A = P^{-1}DP$ where $D$ is upper triangular. What are the diagonal entries of $D$? If $A$ is $n\times n$, what is $A^n$?

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I know that if $A^n$ then the eigenvalues of a nilpotent matrix must be $0$. Is $A = P^{-1}DP$ the only way to prove this? Is there another way other than Jordan Canonical form? – diimension Dec 11 '12 at 4:09
@diimension it's not the only way, see the other two (identical) answers. Also, it doesn't require $D$ to be in JN form, just that it is upper triangular, it is a lot easier to prove that this is possible than JN form. – Tom Oldfield Dec 11 '12 at 4:13
Yes, but I have not learned Cayley-Hamilton theorem yet so I cannot use their proof. – diimension Dec 11 '12 at 4:23
@diimension well then see what you can work out from the hint I gave. I can't think of any other proofs of the top of my head, I'm afraid. – Tom Oldfield Dec 11 '12 at 4:25
@diimension no worries, glad you understand it! – Tom Oldfield Dec 11 '12 at 5:16

If you have learned schur triangularization (or decomposition), note that matrices with all eigenvalues as zero are unitarily similar to "strictly" Upper Triangular matrices. Now see that strictly upper triangular matrices are always nilpotent. Now look at the converse.

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Thank you very much, dineshdileep! – diimension Dec 11 '12 at 5:01

\begin{equation*}\textbf A\vec x=\lambda \vec x\end{equation*} \begin{equation*}\textbf A(\textbf I-\lambda)\vec x=\textbf 0 \end{equation*}

\begin{equation*}(\lambda =0)\land (\textbf I\neq\textbf 0)\land (\vec x\not\equiv \vec0)\implies \end{equation*}

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