# If a Matrix Has Only Zero as an Eigen-Value Then It Is Nilpotent

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

How will I be able to prove this?

• Also, note that the converse is true. Dec 11, 2012 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. Dec 11, 2012 at 4:12

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.

• 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. Dec 11, 2012 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? Dec 11, 2012 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 $\mathbb C$. Dec 11, 2012 at 5:25
• (+1) Can you add some hints on how one should build up a counter example when the field is $\Bbb{R}$? :) Aug 24, 2016 at 21:19
• 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 $0\oplus\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$ Aug 24, 2016 at 22:16

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 $A^n$?

• 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? Dec 11, 2012 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. Dec 11, 2012 at 4:13
• Yes, but I have not learned Cayley-Hamilton theorem yet so I cannot use their proof. Dec 11, 2012 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. Dec 11, 2012 at 4:25
• @diimension no worries, glad you understand it! Dec 11, 2012 at 5:16

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

• I have not learned Cayley-Hamilton theorem yet. Dec 11, 2012 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?

• I have not learned Cayley-Hamilton theorem yet. Dec 11, 2012 at 4:11

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.

• Thank you very much, dineshdileep! Dec 11, 2012 at 5:01