Wrong proof: If $0$ is the only eigenvalue of a linear operator, is the operator nilpotent

Question

I am making the following claim:

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

Proof:

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

Doubt:

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?

Answer 1

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

Answer 2

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, $T^n=0$.