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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. $\endgroup$
    – darij grinberg
    Commented Sep 24, 2021 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). $\endgroup$
    – darij grinberg
    Commented Sep 24, 2021 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. $\endgroup$
    – Richard Stanley
    Commented Sep 24, 2021 at 21:27
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    $\begingroup$ I find the example (complex-valued inner product) not really convincing. Algebraically, a Hermitian form is rather a real object (defined from a quadratic Galois extension, etc). Or, alternatively, one views it defined from a field endowed with an involution (not from a field on which there "exists" an involution). (By the way when one complexifies the real object $(\mathbf{C},$conjugation), one obtains a complex object which is isomorphic to $(\mathbf{C}\times\mathbf{C},$flip). $\endgroup$
    – YCor
    Commented Sep 24, 2021 at 21:44
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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
    Commented Sep 24, 2021 at 23:46

18 Answers 18

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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"? $\endgroup$
    – Steven Landsburg
    Commented Sep 25, 2021 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. $\endgroup$
    – Gerry Myerson
    Commented Sep 25, 2021 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
    Commented Sep 25, 2021 at 11:30
  • $\begingroup$ A finite dimensional vector space is the union of countably many proper subspaces if and only if the underlying field is countable. $\endgroup$
    – Kapil
    Commented Jan 23, 2022 at 10:11
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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. $\endgroup$
    – Andrei Smolensky
    Commented Sep 25, 2021 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
    Commented Sep 25, 2021 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 25, 2021 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
    Commented Sep 26, 2021 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
    Commented Sep 26, 2021 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.") $\endgroup$
    – Sam Hopkins
    Commented Sep 25, 2021 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 $q$. $\endgroup$
    – diracdeltafunk
    Commented Sep 25, 2021 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
    Commented Sep 25, 2021 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
    Commented Sep 26, 2021 at 14:33
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    $\begingroup$ @KConrad What is amazing is that the proportion is never $\frac1q$, even in the limit. $\endgroup$
    – Denis Serre
    Commented Oct 18, 2021 at 9:03
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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? $\endgroup$
    – diracdeltafunk
    Commented Sep 25, 2021 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_
    Commented Sep 25, 2021 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_
    Commented Sep 25, 2021 at 18:21
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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? $\endgroup$
    – André Henriques
    Commented Sep 25, 2021 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) $\endgroup$
    – Tyler Lawson
    Commented Sep 26, 2021 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. $\endgroup$
    – HenrikRüping
    Commented Sep 26, 2021 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. $\endgroup$
    – Michael Hardy
    Commented Sep 27, 2021 at 16:50
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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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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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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 $2 \neq 0$. In a field where $0 = 1 + 1 = 2$, we can confirm that $A = B + C$.

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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
    Commented Sep 26, 2021 at 4:24
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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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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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The subgroup of $\text{GL}_n(k)$ generated by diagonalizable matrices is the whole of $\text{GL}_n(k)$, unless $k=\mathbb{F}_2$ in which case it is trivial.

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    $\begingroup$ There are a number of results on matrix groups over fields that have counterexamples for small finite fields and small matrix sizes. For example, if $n \geq 2$ and $F$ is a field then (i) ${\rm GL}_n(F)' = {\rm SL}_n(F)$ unless $n = 2$ and $|F| = 2$, (ii) ${\rm SL}_n(F)' = {\rm SL}_n(F)$ unless $n = 2$ and $|F| \leq 3$ and (iii) the group ${\rm PSL}_n(F) := {\rm SL}_n(F)/Z$ is simple unless $n = 2$ and $|F| \leq 3$. $\endgroup$
    – KConrad
    Commented Oct 17, 2021 at 20:59
  • $\begingroup$ Yes of course you're right. That particular result just hit my brain when I read HenrikRüping's above post about multilinear forms, because of the fact that the field $\mathbb{F}_2$ has such a singular behaviour (in my example : full group / trivial group). $\endgroup$
    – Matthieu Romagny
    Commented Oct 18, 2021 at 6:16
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This is a very simple fact that besides linear algebra, uses just a bit of geometry/topology. Over the reals, there are pairs of conjugate rotations, say $A$ and $B$, such that $A$ cannot be continuously conjugated into $B$, within $\mathbb R$, but it can do so over $\mathbb C$. For example, take $$ A= \left[ {\begin{array}{cc} c & -s \\ s & c \\ \end{array} } \right] $$ with $c=\cos(\theta)$ and $s=\sin(\theta)$, for some generic real $\theta$, and let $B$ be the transpose of $A$. Then $B$ and $A$ are conjugate via a matrix of negative determinant, so $A$ cannot be continuously conjugated to $B$ unless we work over the complexes.

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    $\begingroup$ Does "$A$ can be continuously conjugated into $B$" mean "there is a continuous path in the conjugacy class of $A$ from $A$ to $B$"? $\endgroup$
    – LSpice
    Commented Jan 12, 2022 at 16:48
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    $\begingroup$ Yes. First, you notice that the conjugacy class of $A$ and $B$ are the same (they are conjugate). Then you find that one can only be conjugated to the other via a matrix of negative determinant. Then, any possible conjugating path, in the real coefficients situation, would have to cross zero determinant. $\endgroup$
    – bernardorim
    Commented Aug 20, 2022 at 16:49
  • $\begingroup$ It may be worth mentioning explicitly that, in addition, $A$ can be conjugated (in $\operatorname{GL}_2(\mathbb R)$) to itself only by a matrix of positive determinant (which can be verified directly, or, I suppose, deduced from the fact that $A$ and $B$ are conjugate, but only by matrices of negative determinant). $\endgroup$
    – LSpice
    Commented Aug 20, 2022 at 16:54
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It is obvious, but easy to forget, that in a field of finite characteristic a non-zero vector can be perpendicular to itself. This has consequences which can be counterintuitive.

For example, a basis for $S$ combined with a basis for $S^\perp$ need not be a basis for the whole space.

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  • $\begingroup$ Is this not also true for the vector $(1, \sqrt{-1}) \in \mathbb{C}^2$? $\endgroup$
    – Martin Seysen
    Commented Jan 23, 2022 at 9:43
  • $\begingroup$ @ Martin Seysen Yes, depending on how you define "perpindicular". In the complex case it's common to conjugate the second vector of the inner product, in which case it's not. My point was that in finite fields the concept of orthogonality is extensively used (to define dual codes, for example) but that the properties of such differ substantially from the real case. $\endgroup$
    – Johnny Calm
    Commented Jan 23, 2022 at 18:12
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The following is true if the field is infinite and false for finite fields.

Let $V$ be a finite-dimensional vector space with subspace $W$. If $f:V\to V$ is linear and $V$ is the smallest $f$-invariant subspace containing $W$, then there exists a $v\in W$ such that the minimal polynomial of $f$ equals the minimal polynomial of $f$ for the vector $v$.

One nice proof of the true "infinite" version uses the answer of Gerry Myerson.

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    $\begingroup$ Does this depend on $g$ itself, rather than just depending on im $g$? If not I feel like it would be more clearly phrased by dropping reference to $g$ and just referring to $U =$ im $g$ as any subspace of $V$ not contained in a proper $f$-invariant subspace. $\endgroup$
    – Bma
    Commented Jan 12, 2022 at 3:00
  • $\begingroup$ @Bma: Good point, thank you. Edited accordingly. $\endgroup$
    – Fabian Wirth
    Commented Jan 12, 2022 at 15:18
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The dimension of $\mathbb{R}$ as an $\mathbb{R}$-vector space is $1$, but $\mathbb{R}$ as a $\mathbb{Q}$-vector space is infinite-dimensional.

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