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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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    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

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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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    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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    @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