2
$\begingroup$

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.

  • 0
    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.)2011-10-01
  • 0
    @Pierre-YvesGaillard Please see my edit above.2011-10-01
  • 0
    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.2011-10-01
  • 0
    @Pierre-YvesGaillard Yes I see what you mean.2011-10-01
  • 0
    @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?2011-10-01
  • 0
    Yes, if $T$ is a nilpotent endomorphism of an $n$-dimensional vector space over a field $K$, then its characteristic polynomial is $X^n\in K[X]$, where $X$ is an indeterminate. Now suppose that $T$ is not necessarily nilpotent, and that $L$ is a field containing $K$. Then the endomorphism $L\otimes_KT$ of $L\otimes_KT$ (if you know this notion) has the same characteristic polynomial as $T$. If $T$ is a matrix (in $M_n(K)$), then its characteristic polynomials in $K[X]$ and $L[X]$ coincide.2011-10-01
  • 0
    @Pierre-YvesGaillard I would like to use the following theorem: Let $T$ be a linear operator on a finite dimensional $K$ - vector space. If the characteristic polynomial of $T$ factors into linear factors $(x-a_1)\ldots (x-a_n)$ with each $a_i \in K$, then there is a basis of the underlying vector space such that $T$ with respect to this basis is upper triangular.2011-10-01
  • 0
    Can you explain to me why your first sentence is true? Let me give it a shot: Every linear operator between on a finite dimensional $K$ - vector space can be represented by a matrix. Its characteristic polynomial, calculated from the determinant will then be a polynomial with coefficients in $K$. Does it then follow that if $0$ is the only complex eigenvalue, the characteristic polynomial must be $X^n$? Maybe you can explain better.2011-10-01
  • 0
    This may not be an intelligent question: Does this mean that any field of characteristic zero is in a way a "subfield" of the complex numbers?2011-10-01
  • 0
    Here's my worry: It is possible for a linear operator over real vector space to have zero as its only eigenvalue, but yet its characteristic polynomial is not $x^n$. This is because $\mathbb{R} \subset \mathbb{C}$. How sure am I that this can't happen when $T$ is nilpotent?2011-10-01
  • 0
    "Can you explain to me why your first sentence is true?": Let's talk in terms of matrices. The eigenvalues are the roots of the characteristic polynomial $\chi$. Now $\chi$ has $n$-roots $a_1,\dots,a_n$ in some field $L$ containing $K$. The $a_i$ are the eigenvalues of your matrix viewed as an endomorphism of $L^n$, and for each $i$ there is an $a_i$-eigenvector, and the nilpotency assumption implies $a_i=0$.2011-10-01
  • 0
    "Does this mean that any field of characteristic zero is in a way a "subfield" of the complex numbers?" No! If $(X)_{i\in I}$ is a family of indeterminates indexed by a set $I$ of cardinality greater than the continuum, then $\mathbb C((X)_{i\in I})$ won't be a subfield of $\mathbb C$.2011-10-01
  • 0
    What do you mean by an $a_i - eigenvector$? Are you speaking of generalised eigenvectors?2011-10-01
  • 0
    "It is possible for a linear operator $T$ over real vector space to have zero as its only eigenvalue, but yet its characteristic polynomial is not $X^n$." Yes. If $n=3$ and $T$ is block diagonal with first block $(0)$ and second block $\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$.2011-10-01
  • 0
    @Pierre-YvesGaillard If $T$ is nilpotent, then every vector in $V$ is a generalised eigenvector of $T$.2011-10-01
  • 0
    "What do you mean by an $a_i$-eigenvector? Are you speaking of generalised eigenvectors?" I do mean eigenvector. - I use the equivalence: $a$ is a root of $\chi$ iff there is a nonzero $v$ such that $Av=av$. I'm taking this for granted.2011-10-01
  • 0
    @Pierre-YvesGaillard What about this: Let $T$ be a linear operator on a $K$-vector space that is nilpotent. This means $T^k = 0$ for some $k > 0$. Now let $m$ be the smallest integer such that $T^m = 0$. Then $T(T^{m-1}(v)) = 0\cdot T^{m-1}v$. Now if $v \neq 0$, then $T^{m-1}(v)$ is not equal to zero for $m$ was assumed to be minimal. So we now know that if $T$ is nilpotent, it has at least zero as an eigenvalue in any field, so we can say: "Let $\lambda$ be an eigenvalue for $T$, so that $Tv = \lambda v$.2011-10-01
  • 0
    Then proceed to show that any such eigenvalue would have to be zero. But then I come to the same problem again on how to conclude about the characteristic polynomial....2011-10-01
  • 0
    Dear D B Lim, I tried to explain this in the comment starting with $<<$ "Can you explain to me why your first sentence is true?": Let's talk in terms of matrices. $>>$ If you disagree with this comment, please say why.2011-10-01
  • 0
    I get it: the characteristic polynomial $\chi$ has roots $a_1 \ldots a_n$ that lie in some larger field $L$ containing $K$. This means that $\chi = (x - a_1)\ldots (x-a_n)$. Now $T$ will have eigenvectors if we consider the larger field $L$, so we can speak of them. Proceed as usual to write down $Tv = av$. We can now write this down if we consider the larger field. But then the nilpotency bit means that any eigenvalue is zero, and so consequently substituting each $a_i = 0$ into the characteristic polynomial shows that $\chi = x^n$.2011-10-01
  • 0
    Exactly!!! ${}$2011-10-01
  • 0
    Now it only is left to disect what we mean by larger field......2011-10-01
  • 0
    Yes, it's one of the mathematical miracles! If you have a non constant polynomial $f$ in $K[X]$ which has no roots in $K$, then you can artificially concoct a larger field $L$ in which $f$ has at least one root. Here is the recipe. You can assume $f$ is irreducible, and put $L:=K[X]/(f)$. Then you can iterate this.2011-10-01
  • 0
    @Pierre-YvesGaillard Je voudrais vous dire que je suis très ravi et reconnaissant que vous m'avez patiemment expliqué les maths.2011-10-01
  • 0
    Do you have a reference where I can read about this? Life's getting more amazing now for me.2011-10-01
  • 0
    Tout le plaisir a été pour moi! Une référence standard: **Algebra** par Michael Artin. (Il y en a beaucoup d'autres.)2011-10-01
  • 0
    Par hasard c'est le texte que l'on utilise pour le cours algèbre I... J'ai la deuxième édition du texte - je crois que je peux trouver les faits que vous avez dit dans chapitre 15 - le corps commutatif (fields).2011-10-01
  • 0
    J'ai la première édition. Le titre du chapitre 13 est *Fields*. Le titre du paragraphe ("section") 5 est *Symbolic Adjunction of Roots*. C'est bien ce que j'ai évoqué.2011-10-01

3 Answers 3

2

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 $v\neq0$).

1

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.

  • 0
    By the same zero I am not referring to the characteristic of the field, but the same additive identity as $\mathbb{C}$.2011-10-01
  • 0
    What does that mean?2011-10-01
  • 0
    I mean in that field, its additive identity is just the number zero.2011-10-01
  • 0
    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.......2011-10-01
  • 0
    I see you changed your question. Let me edit my answer accordingly.2011-10-01
  • 0
    Anyway, if you say "same additive identity as $\mathbf{C}$" the only thing I can make out of this is "characteristic zero".2011-10-01
  • 0
    Over the complex numbers it is easy to see the only eigenvalues of $T$ are zero. Is it true that over any field, the eigenvalues of a nilpotent operator are all zero?2011-10-01
  • 0
    You can do it over the complex numbers without jordan form. Just note that there is a basis for which the matrix with respect to the basis of T is upper triangular.2011-10-01
  • 0
    If $Tx = ax$ and $T^n =0$,we have that $a^n =0$. Therefore, $a=0$. This is independent of the characteristic of the field. Thus the eigenvalues of a nilpotent operator are all zero. When you say "just note that there is a basis in which the matrix is upper triangular" you are using a special case of the Jordan form. If I'm not mistaken, to prove the Jordan normal form theorem you use this "special case".2011-10-01
  • 0
    @Gooz: The Jordan normal form theorem doesn't hold over any field. Look for the sentences "If the ground field is not perfect, then a Jordan–Chevalley decomposition may not exist. Example: ..." in this Wikipedia [entry](http://en.wikipedia.org/wiki/Jordan–Chevalley_decomposition). [The link doesn't seem to work. The title of the entry is "Jordan–Chevalley decomposition".]2011-10-01
  • 0
    @shaye: Please see my previous comment. (Sorry, I can notify only one user at a time.)2011-10-01
  • 0
    @shaye I know about that result, its just the field in question that is confusing. Now what can we say about the characteristic polynomial of $T$ if its only complex eigenvalues are zero? Is the characteristic polynomial then just $x^n$? Do I need to speak of a field first before even talking of the characteristic polynomial?2011-10-01
  • 0
    The characteristic polynomial of a linear transformation $f:V\to V$ makes sense over any field (and even comm. ring). It's the same definition: $p_f(x) =det (f-x\cdot I)$. If $f$ is nilpotent, then the characteristic polynomial is $p_f(x) = x^n$, where $n = \dim V$. One can also show that the converse is true in characteristic zero. I'm not sure if this is true in any characteristic. That is, $f$ is nilpotent iff $p_f(x) = x^n$.2011-10-01
1

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.)

  • 0
    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).2011-10-01