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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.2012-12-11
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    You need to make assumptions on the base field for this to work. It won't work over $\Bbb R$ for instance.2012-12-11

5 Answers 5

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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 $A^n=(VTV^*)^n=VT^nV^*=0$, and $A$ 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.2012-12-11
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    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?2012-12-11
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    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 $\mathbb C$.2012-12-11
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    (+1) Can you add some hints on how one should build up a counter example when the field is $\Bbb{R}$? :)2016-08-24
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    Not sure what you mean: there is already a counter-example in my answer.2016-08-24
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    I know there is a counter example but I want to know how did you make that counter example! :) There is always some thought behind making counter examples, isn't it? :)2016-08-24
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    Oh, ok. The canonical example of a matrix with no real eigenvalues is $$\begin{bmatrix}0&1\\-1&0\end{bmatrix},$$ because its characteristic polynomial is $t^2+1$ (the easiest polynomial with non-real roots). Then I enlarged the example to have a real eigenvalue (zero), so I see the matrix in the example as a direct sum $0\oplus\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$2016-08-24
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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?2012-12-11
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    @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.2012-12-11
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    Yes, but I have not learned Cayley-Hamilton theorem yet so I cannot use their proof.2012-12-11
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    @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.2012-12-11
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    Okay then, thanks for the insight!2012-12-11
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    @diimension let me know if you need a stronger hint.2012-12-11
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    Yes, please do ..2012-12-11
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    @diimension Okay, well $A^n = P^{-1}D^nP$ from above. If the diagonal entries of $D$ are all $0$ (which they must be, if the eigenvalues of $A$ are all zero), what is $D^n$?2012-12-11
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    If D is strictly upper triangular matrix then $D^n$ will also be strictly upper diagonal matrix?2012-12-11
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    @diimension that's true, but we can say more than that. Try some examples.2012-12-11
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    Should I have added that fact that $D^n$ will have $0$ eigenvalues? Will that suffice?2012-12-11
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    @diimension just try some example $D^n$s when $n = 2$ or $3$. It should make a lot more sense2012-12-11
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    K, can you define what P is?2012-12-11
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    Actually, I got it thank you very much!2012-12-11
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    @diimension no worries, glad you understand it!2012-12-11
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Hint: Look at the characteristic polynomial, then use Cayley-Hamilton.

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    I have not learned Cayley-Hamilton theorem yet.2012-12-11
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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.2012-12-11
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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!2012-12-11