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Linear algebra as we learn it as undergraduates usually holds for any field (even though we usually learn it for the complex, or real, numbers).

I am looking for a list of concepts, and results, in linear algebra that actually depend on the choice of field.

To start I propose the notion of an complex valued inner product. Here the anti-linear axiom requires an involution on the field.

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    $\begingroup$ The famous $\operatorname{Ker} A = \operatorname{Ker}\left(A^T A\right)$ (which is used, e.g., in the construction of the Moore-Penrose pseudoinverse) requires the field to be ordered. $\endgroup$ Sep 24 at 20:46
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    $\begingroup$ The theory of skew-symmetric matrices depends on the field unless one is careful and studies alternating matrices instead (as one should 95% of the time). $\endgroup$ Sep 24 at 20:47
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    $\begingroup$ Solvability of $AB-BA=I$ for $n\times n$ matrices $A,B$ depends on the field characteristic. Also diagonalizability of the $n\times n$ all 1's matrix. $\endgroup$ Sep 24 at 21:27
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    $\begingroup$ The premise of this question is incorrect because the part of such a course involving inner products does not work over general fields (positive-definiteness as opposed to mere nondegeneracy of a symmetric bilinear form, square roots of inner products, and the spectral theorem). $\endgroup$
    – KConrad
    Sep 24 at 23:46
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    $\begingroup$ @JochenGlueck, I think it matters quite a bit where and for whom you teach. I teach at a small US university where the linear-algebra course serves many non-math majors, and most of the course is taught over just the real numbers, with even the complex numbers mentioned only when necessary. $\endgroup$
    – LSpice
    Sep 25 at 14:14

13 Answers 13

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The existence of Chevalley–Jordan decompositions depends on the perfectness of the field.

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A finite-dimensional vector space is a union of finitely many proper subspaces if and only if the underlying field is finite.

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    $\begingroup$ Or how about just "a finite dimensional vector space is finite"? $\endgroup$ Sep 25 at 3:59
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    $\begingroup$ @Steven, it's true that one direction of the "if and only if" is trivial (although, come to think of it, I should insist on the dimension being at least two), but the other direction requires an argument. $\endgroup$ Sep 25 at 6:04
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    $\begingroup$ I would prefer to replace vector spaces by affine spaces in this proposition. The version for affine spaces follows more directly from induction (passing to hyperplanes). $\endgroup$
    – Z. M
    Sep 25 at 11:30
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As mentioned in the comments: when the characteristic of your field is not $2$, "skew-symmetric" and "alternating" are equivalent conditions on a bilinear form. In characteristic $2$, alternating implies skew-symmetric (which is also equivalent to symmetric), but not vice versa.

For example, multiplication as a bilinear form on $\mathbb{F}_2$ is (skew-)symmetric, but not alternating since $1 \cdot 1 = 1 \neq 0$.

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Existence of Jordan canonical form (requires algebraically closed field).

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    $\begingroup$ Not really. The theorem should be stated in the form "if the characteristic polynomial splits over the base field, the operator admits a Jordan form", the requirement of the field being closed is there only to force the assumption. $\endgroup$ Sep 25 at 14:58
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    $\begingroup$ @Andrei Smolensky: yes, sure. But maybe Brauer Suzuki by that phrase was meaning "existence of Jordan canonical form [for all operators]" $\endgroup$
    – Qfwfq
    Sep 25 at 19:17
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    $\begingroup$ @AndreiSmolensky, well, add the clause "for all endomorphisms of all finite dimensional vector spaces" if that feels more satisfying ¯_(ツ)_/¯ You can almost always replace the hypotehsis that a field is algebraically closed by one demanding that certain specific polynomials split. One generally does not. $\endgroup$ Sep 25 at 21:48
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    $\begingroup$ Existence of eigevalues "for all endomorphisms of all finite dimensional vector spaces" requires algebraically closed fields. $\endgroup$
    – Nick S
    Sep 26 at 4:12
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    $\begingroup$ But there are natural generalizations which work over any field, such as the "primary rational canonical form". These are given in terms of companion matrices of irreducible polynomials. Of course in the algebraically closed case this just gives $1 \times 1$ matrices $(\lambda)$, corresponding to polynomial $x - \lambda$. $\endgroup$
    – spin
    Sep 26 at 12:34
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For a finite field ${\mathbb F}_q$, you may calculate the probability that the determinant of an $n\times n$ matrix is $0$. This probability has a limit $\pi_q$ as $n\rightarrow+\infty$. Amazingly, this $\pi_q$ does depend upon $q$. In particular, it is $>\frac1q$.

To be more precise, the probability that $\det M\ne0$ is, for fixed $n$, $$\prod_{m=1}^n\left(1-\frac1{q^m}\right).$$ Its limit as $n\rightarrow+\infty$ is non trivial and is strictly less than the first factor $1-\frac1q$. Hence $$\pi_q=1-\prod_{m=1}^\infty\left(1-\frac1{q^m}\right)>\frac1q.$$ Notice that this can be expressed in terms of Dedekind's eta function.

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    $\begingroup$ Implicitly you're using the uniform distribution on the set of such matrices here, right? I wonder if there is some other "natural" distribution where the answer does not depend on $q$. (I also don't totally see why the dependence on $q$, for the uniform distribution, should be "amazing.") $\endgroup$ Sep 25 at 14:58
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    $\begingroup$ Could we list some example values of $\pi_q$? The bounds $1/q < \pi_q \leq 5/8$ do not immediately preclude the possibility that $\pi_q$ is independent of $q$. $\endgroup$ Sep 25 at 17:24
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    $\begingroup$ @diracdeltafunk The log of the product in question gives an absolutely convergent series in $q$ which is increasing in $q$. $\endgroup$
    – JoshuaZ
    Sep 25 at 17:59
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    $\begingroup$ Interesting, but I'm not sure it's very suprising. It's certainly not for $n=1$ :-) $\endgroup$
    – leonbloy
    Sep 26 at 14:33
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The trueness of the statement "two vector spaces are isomorphic if and only if their dual spaces are isomorphic" depends on the cardinality of the field (and the underlying set-theoretic axioms).

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  • $\begingroup$ What is this dependence? $\endgroup$ Sep 25 at 17:26
  • $\begingroup$ @diracdeltafunk - Let me give two examples. But first note, if we accept the axiom of choice, then, for an infinite dimensional vector space $V$ over a field $k$ we have $\dim(V^\ast)=|k|^{\dim(V)}$. Example 1: Let $|k|=|2^\mathbb{R}|$ and $\dim(V)=|\mathbb{N}|, \dim(W)=|\mathbb{R}|$. Hence $V^\ast,W^\ast$ are isomorphic because they have the same dimension $|k|$. So, for $k$ the statement doesn't hold. $\endgroup$
    – tj_
    Sep 25 at 18:20
  • $\begingroup$ Example 2: Let $k$ be a countable field and $V,W$ be infinite dimensional vector spaces over $k$ such that $V^\ast\cong W^\ast$. Hence $|2^{\dim(V)}|=\dim(V^\ast)=\dim(W^\ast)=|2^{\dim(W)}|$ and by the generalized continuum hypothesis $\dim(V)=\dim(W)$ follows. So, for $k$ the statement is true. $\endgroup$
    – tj_
    Sep 25 at 18:21
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A symmetric tensor is a linear combination of tensor powers over a field of characteristic 0 (or large enough), but not always.

(The underlying reason is that polarization formulae contain denominators.)

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Let $U_n$ be a $n \times n$ Jordan block with $1$'s on the diagonal (unipotent Jordan block).

Then for $n,m > 0$ the Kronecker product $U_n \otimes U_m$ has a Jordan normal form over any field, but the Jordan blocks that occur depend on the characteristic of the field.

There is a closed formula in characteristic $p = 0$ and $p \geq m+n$, in which case $U_n \otimes U_m$ is similar to the matrix $$U_{n+m-1} \oplus U_{n+m-3} \oplus \cdots \oplus U_{n+m-2s+1}$$ where $s = \min(m,n)$.

But in general there is no such formula (except recursive ones). For example in characteristic $p > 0$ you get $$U_p \otimes U_p \sim U_p \oplus \cdots \oplus U_p\ (p \text{ times})$$

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The vector space of multilinear maps $\prod_{i=0}^\infty\mathbb{F}\rightarrow \mathbb{F}$ is infinite dimensional, unless the field is $\mathbb{F}_2$, in which case it is one dimensional.

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    $\begingroup$ This exercise is a bit difficult. Do you mind including a hint? $\endgroup$ Sep 25 at 20:43
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    $\begingroup$ @AndréHenriques The vector space $$colim_n (V_1 \otimes \dots \otimes V_n) \otimes \Bbb F\{\prod_{k>n} V_k\}$$ has the right universal property to accept a universal multilinear map from $\prod V_k$, and the colimit description makes it easier to check when two elements are equal (roughly, only when there is a simple reason so) $\endgroup$ Sep 26 at 4:02
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    $\begingroup$ yes and then one can replace $\prod_{k>n}V_k$ by $\prod_{n>k}(V_k\setminus 0)$ to see that it is one-dimensional for $V_k=\mathbb{F}_2$. one can also write down a basis for the tensor product given by a choice of representatitives of $(\prod_n (V_k\setminus 0))/\sim$ where two sequences are equivalent, iff they differ only in finitely many places. $\endgroup$ Sep 26 at 8:44
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    $\begingroup$ @HenrikRüping : Notice this typographical difference: $$(\prod_n (V_k\setminus 0))/\sim$$ $$(\prod_n (V_k\setminus 0))/{\sim}$$ The $\text{“}{\sim}\text{”}$ symbol has horizontal space between it and whatever is to its left or right because it's a binary relation symbol, except when nothing is to its left or right. Since that spacing is not appropriate in the expression above, you can code it as (\prod_n (V_k\setminus 0))/{\sim} with {curly braces} enclosing it. That's the way to do it. $\endgroup$ Sep 27 at 16:50
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In representation theory (a bit beyond pure linear algebra):

Maschke's theorem: a finite-dimensional representation of a finite group $G$ over a field $k$ with characteristic not dividing the order of $G$ is semisimple.

Weyl's theorem on complete reducibility: every finite-dimensional representation of a semisimple Lie algebra over a field of characteristic zero is semisimple.

So in characteristic zero it is always semisimple, but in positive characteristic, not always.

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One such property i had in an exam once was this one:

Are $A, B, C$ linear independent vectors?

$$ A = \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}, \ B = \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \ C = \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix} $$ We defined one way to tests that property over the determinant:

$ \text{det}(\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{bmatrix})$

which is $2$. So are they linearly independent? Only in a field where $2 \neq 0$, a field where $0 = 1 + 1 = 2$ is the boolean algebra. We can confirm that via $A = B + C$.

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The existence of two commuting nilpotent matrices of specified Jordan types depends on the characteristic and size of the field. This was first shown in my joint paper with John Britnell, Types and classes of commuting matrices, J. Lond. Math. Soc. 83 (2011) 470–492.

Specifically, Proposition 4.7 states that there are matrices of Jordan types $(n,n)$ and $(n+1,n-1)$ that commute over $\mathbb{F}_{p^r}$ if and only if $n$ is not divisible by $p(p^{2r}-1)/e$, where

$$e = \begin{cases} 1 & \text{if $p=2$} \\ 2 & \text{otherwise.} \end{cases}$$

The smallest example of this type is that there are matrices of Jordan types $(6,6)$ and $(7,5)$ that commute with entries in $\mathbb{F}_{4}$, $\mathbb{F}_8$, and so on, but no such matrices with entries in $\mathbb{F}_2$.

Section 4.4 of the paper gives some further field-dependent results of this type, including a classification of all commuting Jordan types labelled by partitions with at most two parts.

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Copied into an answer from a comment of D. Grinberg:

The famous $\ker A=\ker(A^T A)$ (which is used, e.g., in the construction of the Moore-Penrose pseudoinverse) requires the field to be ordered.

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    $\begingroup$ I think that the Moore-Penrose pseudoinverse is based on $\ker(A)=\ker(A^*A)$ which is true over $\mathbb C$. $\endgroup$
    – Nick S
    Sep 26 at 4:24

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