I am making the following claim:

If $0$ is the only eigenvalue of a linear operator $T$, then $T$ is nilpotent


Since $0$ is the only eigen value, the charecteristic polynomial is of the form $p(x)=x^n$. Then $T^n=0$.


But this answer suggests that we need to assume that the underlying field is algebraically closed.

I cannot understand what am I doing wrong. Any idea?

  • 1
    $\begingroup$ If you say : $0$ is the only real eigenvalue, then the above statement is incorrect. If you say $0$ is the only complex eigenvalue, then you are right. The difference between real and complex numbers lies in the fact that the latter is algebraically closed while the former is not. So even if the characteristic equation admits only the real eigenvalue zero, it may admit other complex eigenvalues that will ensure that the characteristic polynomial is not of the form above. An example is $x(x^2+1)$ of this phenomena. $\endgroup$ Jan 22 '18 at 1:48

It's hard to give an answer without more or less saying what was said in the other answer.

The matrix $\begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix}$ has no real eigenvalues. To see this, it's characteristic polynomial is $x^2+1$, which has no real roots. Thus every real eigenvalue is vacuously $0$, but the matrix is not nilpotent.

In this case there are complex eigenvalues which are nonzero, but there are no real eigenvectors corresponding to these.

However, if we were in an algebraically closed field such as $\mathbb{C}$, and $T$ had no eigenvalues, then since the characteristic polynomial splits into linear factors, i.e. is of the form $\Pi(x-\alpha_i)$, each of the $\alpha_i$ is an eigenvalue, so if they are all $0$, then the characteristic polynomial, as you say, must be $x^n$ so by Cayley-Hamilton, $T^n=0$.


If the field isn't algebraically closed, $p(x)$ might not be $x^n$; for example, over $\mathbb{R}$ it might be $x(x^2+1)$.


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