Results in linear algebra that depend on the choice of field 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.
 A: 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.

A: 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.
A: 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.
A: 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.
A: The existence of Chevalley–Jordan decompositions depends on the perfectness of the field.
A: 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.
A: A finite-dimensional vector space is a union of finitely many proper subspaces if and only if the underlying field is finite.
A: 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 $1 \cdot 1 = 1 \neq 0$.
A: Existence of Jordan canonical form (requires algebraically closed field).
A: 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.
A: 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).
A: 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 $$U_p \otimes U_p \sim U_p \oplus \cdots \oplus U_p\  (p \text{ times})$$
A: 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.
A: 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.)
A: 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 $A = B + C$.
A: 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.
A: 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.
A: 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.
