If the field of a vector space weren't characteristic zero, then what would change in the theory? In the book of Linear Algebra by Werner Greub, whenever we choose a field for our vector spaces, we always choose an arbitrary field $F$ of characteristic zero, but to understand the importance of the this property, I am wondering what would we lose if the field weren't characteristic zero ?
I mean, right now I'm in the middle of the Chapter 4, and up to now we have used the fact that the field is characteristic zero once in a single proof, so as main theorems and properties, if the field weren't characteristic zero, what we would we lose ?
Note, I'm asking this particular question to understand the importance and the place of this fact in the subject, so if you have any other idea to convey this, I'm also OK with that.
Note: Since this a broad question, it is unlikely that one person will cover all the cases, so I will not accept any answer so that you can always post answers.
 A: When doing basic linear algebra, there is no real advantage for the theory in assuming a field of characteristic zero. (Nor, I should add, is there any real advantage in assuming commutativity: until doing eigenvalue problems, working over a division ring is perfectly fine. Indeed not assuming commutativity is a very good exercise in mental discipline, keeping scalars to one side and matrices to the other.)
There is a practical advantage that in examples one can write down explicit scalars that are obviously unequal; without any assumption on the characteristic, any integer except $-1,1$ might fail to be nonzero, and beginning students might be surprised e.g. that $\frac{13}{16}=\frac9{14}$ when the characteristic is $19$.
A: When dealing with inner products, you need to use inequalities, so you have to work with an ordered field (in general $\Bbb R$). Thus anything that is proved using to inner products need not be true in a field of positive characteristic; for example, a symmetric matrix over a finite field is not necessarily diagonalisable. For example, in a field of characteristic $2$ the matrix
$$\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$$
is nilpotent, but not zero, and thus it is not diagonalizable.
In fact, your previous question is another example (perhaps that's the case you mention in your question): it doesn't hold in characteristic $2$, because in the proof you need to divide by $2$. For example, the bilinear form
$$\phi :\Bbb F_2^2\times \Bbb F^2_2\to \Bbb F_2:((x_1,x_2),(y_1,y_2))\mapsto x_1y_1+x_2y_2$$
is skew-symmetric in the sense that $\phi(\vec{x},\vec{y})=\phi(\vec{y},\vec{x})=-\phi(\vec{y},\vec{x})$, but $\phi((1,0),(1,0))\neq 0$.
A: Two problems that I have myself come across are:


*

*The invertibility of a matrix can change. For example,
the integer matrix $$\left[\begin{matrix}4 & 1\\2 & 2\end{matrix}\right]$$ is invertible over the real field because its determinant is nonzero (it is six). Considering the same matrix over the field $\mathbb{F}_3$ (with characteristic three), the determinant is zero and the same matrix is now singular. In this case the matrix should be written as $$\left[\begin{matrix}1 & 1\\2 & 2\end{matrix}\right]$$ and it is easy to see that one row is the multiple of the other.

*This one is the shocking one that you can have a nonzero vector with norm zero. For example, consider $$x=[1,1,1]$$ which is obviously a nonzero vector over the field $\mathbb{F}_3$ but using the usual inner-product and norm definitions $$||x||^2=|<x,x>|^2=0.$$ So this nonzero vector is orthogonal to itself. Over the real and the complex field, only the zero vector is orthogonal to itself. Now just see how many of the standard linear algebra algorithms fail because of this fact. For one, QR decomposition won't work because Gram-Scmidt won't work here. Something as simple as normalizing a vector would fail.

A: "Skew-symmetric" is different from "alternating" in characteristic $2$, but this is more of a multilinear (or exterior) algebra issue than a linear algebra issue, and it's more of a "why characteristic $2$ matters" example than a "why characteristic $0$ matters" one. But it gives a flavor of what can go wrong in positive characteristic: you might not be able to divide by a constant you'd like to divide by. This is the same problem that arises in lhf's example of the polarization identity.
(More precisely, we have alternating $\implies$ skew-symmetry in all characteristics, but the converse holds only for characteristic not equal to $2$.)
A: Many arguments using the trace of a matrix will no longer be true in general. For example, a matrix $A\in M_n(K)$ over a field of characteristic zero is nilpotent, i.e., satisfies $A^n=0$, if and only if $\operatorname{tr}(A^k)=0$ for all $1\le k\le n$. For fields of prime characteristic $p$ with $p\mid n$ however, this fails. For example, the identity matrix $A=I_n$ then satisfies  $\operatorname{tr}(A^k)=0$ for all $1\le k\le n$, but is not nilpotent. 
The pathology of linear algebra over fields of characteristic $2$ has been discussed already here.
A: The equivalence between symmetric bilinear forms and quadratic forms given by the polarization identity breaks down in characteristic $2$.
A: One important thing is there is no geometry of physical space when we work over finite fields. The interpretation for $\det A$ as the scaling factor of the 
"volume"  is not available any more in finite fields. This is for the simple reason that elements of real field are measurement of quantities (numbers) whereas elements of finite field are not actually numbers, they don't measure any quantities, they happen to satisfy all the abstract axioms of a field.
So rotations, reflections etc which are geometric linear transformations do not have any intuitive explanation over finite fields. In a field with $q$ elements, for any eigenvector $v$ of any matrix $A$, we have $A^{q-1}v=v$.
In real field $A^{q-1} v$ would be a vector for away from $v$ (if the eigenvalue is of magnitude > 1).
A: Nowhere is the difference between characteristic $0$ and characteristic $p>0$ as stark as in the representation theory of finite groups.
Let me try to illustrate this with a baby example:
Let $V$ be a vector space together with a linear map $f\colon V\to V$ such that $f^p=1$ (where $1$ denotes the identity on $V$). This is precisely a representation of the cyclic group $\mathbb Z/p\mathbb Z$ but you can safely ignore that if you don't know about groups.
Assume we have a subspace $W\subset V$ which is stable under $f$, in the sense that $f(w)\in W$ for all $w\in W$. Then we could ask, if we can find a complement $X$ of $W$ in $V$ which is also stable under $f$. In other words, we want to know, whether we can "break up" $V=X\oplus W$ such that $f$ "acts separately on the two parts", i.e. we can write $f(x,w)=(f(x),f(w))$.
It turns out that you can always do this in characteristic zero! (you can look this up under "Maschke
's theorem"; don't let the name scare you, it is not a difficult proof)
But consider for instance the puny $2$-dimensional vector space $V= F^2$, where $F=\mathbb F_p$ is the field with $p$ elements. Consider $f$ given by the matrix $\left(\begin{matrix}1& 1\\ 0 &1\end{matrix}\right)$. Clearly $f^p=1$ (because $p\cdot 1=1+1+\dots+1=0$ in $F$). Let $W$ be subspace of $W$ spanned by the first basis vector $e_1=\left(\begin{matrix}1\\ 0\end{matrix}\right)$; clearly $W$ is $f$-stable because $f(e_1)=e_1$. But a complement of $W$ would have to be spanned by a vector of the form $x=\left(\begin{matrix}a\\ b\end{matrix}\right)$ with $b\neq 0$; hence $f(x)=\left(\begin{matrix}a+b\\ b\end{matrix}\right)$ is not a multiple of $x$. So $W$ has no $f$-stable complement.
It turns out that this seemingly small issue has enormous ripples. For instance, the representation theory of the symmetric group is extremely well understood in characteristic zero (you can find everything you need to know in any introductory textbook); on the other hand, understanding the representation theory of the symmetric group in positive characteristic  is one of the big open problems of modern mathematics.
A: One important difference (which I don't see in any other answer) is that in fields of non-zero characteristic, we can't have a "norm" or "inner product" the way we might over $\Bbb R, \Bbb C,$ or even $\Bbb Q$.  In particular: in order to make sense of conditions like
$$
\|\alpha x\| = |\alpha| \cdot \|x\|\\
\langle x,x \rangle \geq 0
$$
It is important to have a notion of "positive numbers" (i.e. we must have an ordered subfield) which we lack for non-zero characteristics.
