# Eigenvalues of a linear operator over a K-vector space

The following question may be somewhat ill-posed due to a lack of experience in dealing with fields.

Let $T$ be a linear operator over a finite dimensional complex vector space. If the only (complex) eigenvalues of $T$ are zero, then does this mean that the only eigenvalues of $T$ in any subfield of $\mathbb{C}$ are zero too? The converse of this is definitely not true.

More generally, for any field $K$ whose additive identity is the same "zero" as that of $\mathbb{C}$, are the only eigenvalues of $T$ in such a field 0?

I am asking this question for I was posed a question about linear operators over a vector space whose field over which was not specified.

$\textbf{Edit :}$ Perhaps I should provide context to what I am asking. I would like to show that if an operator on a vector space over some field $K$ is nilpotent, then the operator in some basis of the vector space is upper triangular with all zeros on the diagonal.

If the vector space is complex this is trivial, and over a general field I know how to prove this without talking about eigenvalues and characteristic polynomials (by noting that one has the ascending chain of nullspaces of powers of $T$ that eventually equals the whole space).

I have added the homework tag too because the problem I am asking now is related to a homework problem. Namely, the one on nilpotent operators I have mentioned above.

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Let $K$ be a subfield of a field $L$, let $X$ be an indeterminate, and let $f\in K[X]$ be nonzero. Then, in general, some of the roots of $f$ in $L$ will lie in $K$, and the others will not. (This applies in particular to the characteristic polynomial of a matrix.) –  Pierre-Yves Gaillard Oct 1 '11 at 9:54
@Pierre-YvesGaillard Please see my edit above. –  user38268 Oct 1 '11 at 11:16
I'll tell you how I understand your edit, and you'll tell me if I'm right. You want a proof of the fact below which uses explicitly the notions of eigenvalue and characteristic polynomial. Fact: if an operator on a (finite dimensional) vector space over some field $K$ is nilpotent, then the operator in some basis of the vector space is upper triangular with all zeros on the diagonal. –  Pierre-Yves Gaillard Oct 1 '11 at 11:29
@Pierre-YvesGaillard Yes I see what you mean. –  user38268 Oct 1 '11 at 11:33
@Pierre-YvesGaillard If $T$ is nilpotent, and its only complex eigenvalues are zero, can we say that its characteristic polynomial is $T^n$? Or to even talk about the characteristic polynomial we need to first specify the field? –  user38268 Oct 1 '11 at 12:34

I convinced myself that the notion underlying the question was that of splitting field for a polynomial $P\in K[X]$, where $K$ is a field and $X$ an indeterminate. In the question, $P$ is the characteristic polynomial of an endomorphism, but this is not (I think) the crucial point.

The notion of splitting field is described in many textbooks and handouts. I will only give a few pointers to online material on the subject.

$\bullet\$ The Wikipedia entry Splitting Fields.

$\bullet\$ A classic reference: Galois Theory: Lectures Delivered at the University of Notre Dame, by Emil Artin. See especially Section D, entitled Splitting Fields, in Chapter II, entitled Field Theory.

$\bullet\$ Chapter II, entitled Splitting fields; multiple roots, of Fields and Galois Theory. This belongs to J.S. Milne's Mathematics Site.

Let me address now one of the sub-questions. Show the following fact:

(I) Let $V$ be an $n$-dimensional vector space over a field $K$, let $f$ be an endomorphism of $V$, and let $P_f\in K[X]$ be the characteristic polynomial of $f$. Then $f^n=0$ if and only if $P_f=X^n$.

The goal is of course to give (or at least to sketch) a proof as elementary as possible.

Statement (I) results immediately (by induction on $n$) from Statements (II) and (III) below, which I'll consider either as known from the reader, or as easy to check.

(II) The following conditions are equivalent:

(a) $\ker f\neq0$,

(b) $X$ divides $P_f$,

(c) there is a basis of $V$ relative to which the matrix of $f$ has the form $$\begin{pmatrix}A& 0\\ C&0\end{pmatrix},$$ where $A$ is an $n-1$ by $n-1$ matrix.

(III) If (c) holds, then $P_f=XP_A$, where $P_A$ is the characteristic polynomial of $A$. Moreover, we have, for all $k\ge1$, $$\begin{pmatrix}A& 0\\ C&0\end{pmatrix}^k=\begin{pmatrix}A^k& 0\\ CA^{k-1}&0\end{pmatrix}.$$

Basically, the only thing the above argument assumes as known is the two equivalent definitions of an eigenvalue (via the equations $P_f(\lambda)=0$ or $fv=\lambda v$ with $v\neq0$).

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I'm not sure if I understand your question correctly but let me try to answer anyway.

If you have a linear operator $T$ on a complex vector space $V$ such that the eigenvalue equation $Tx = \lambda x$ only has the solution $\lambda =0$ over $\mathbf{C}$, you can conclude that the only eigenvalues of $T$ are zero. In fact, the solutions of this equation are the zeroes of the polynomial $\det(T-\lambda \cdot I)$ (which has coefficients in $\mathbf{C}$).

So to be clear, for any subfield $k$ of $\mathbf{C}$ you have that the eigenvalues of $T$ lying in $k$ are zero simply because they lie in $\mathbf{C}$.

The rest of your question I don't quite understand. What do you mean by "the same "zero" as that of $\mathbf{C}$. Maybe you mean a field extension of $\mathbf{C}$? Like the field of rational functions over $\mathbf{C}$? Or do you mean any field of characteristic zero?

In general, if you have a linear operator on a $k$-vector space $V$, where $k$ is any field, the eigenvalues of $T$ all lie in an algebraic closure of $k$. This is simply because the eigenvalues are the zeroes of the polynomial $\det(T-\lambda \cdot I)$ which has coefficients in $k$.

Hope that helps.

Edit: The OP added a question.

Firstly, it is not trivial that over the complex numbers $T$ takes this form. This is just a special case of the Jordan normal form theorem. Anyway, the proof of the Jordan normal form theorem works over any algebraically closed field. If you are willing to accept this the answer to your question is easy.

The eigenvalues of a nilpotent linear operator are all zero. Thus, the Jordan normal form is an upper triangular matrix with zeroes on the diagonal.

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By the same zero I am not referring to the characteristic of the field, but the same additive identity as $\mathbb{C}$. –  user38268 Oct 1 '11 at 10:49
What does that mean? –  Gooz Oct 1 '11 at 11:01
I mean in that field, its additive identity is just the number zero. –  user38268 Oct 1 '11 at 11:07
The additive identity of a field (or even ring) is called the zero element. There is a unique morphism from the ring of integers to any field. That is, if $k$ is a field then there is a unique morphism $\mathbf{Z} \to k$. This morphism sends the NUMBER zero to the additive identity of $k$. So I guess all fields have the number zero as an additive identity....... –  Gooz Oct 1 '11 at 11:31
I see you changed your question. Let me edit my answer accordingly. –  Gooz Oct 1 '11 at 11:33

The trick is to use the rational normal form (aka Frobenius normal form). If your matrix is defined over $K$ then in compouting this form, all required calculations are performed in $K$ and so the result is a matrix with coefficients in $K$. The blocks of the normal form are the "companion matrices" of polynomials and the matrix is nilpotent if and only if each block is triangular. (I am leaving out some details, but the question is labelled as homework.)

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We do not learn about Frobenius Normal Forms, so even if you tell me the whole proof it will not be any use (I cannot use it in an assignment). –  user38268 Oct 1 '11 at 13:40