TITLE: Elementary linear algebra over a (possibly skew) field $K$
QUESTION [5 upvotes]: I have a number of questions which seem linked to me, about basic  (?) linear algebra:
Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $L$ (any reference for basic facts about that ?)
So given a square matrix $M\in M_{n\times n}(L)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $L^n$ and its right kernel ? 
By left kernel, I mean the space (left-vector space) of rows $\lambda\in L_{1\times n}$ (one row $n$ columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in L_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.
Question 1. For instance, given $M\in M_{n\times n}(L)$, if there is a non-trivial point  of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?
If $K$ is commutative, I am trying to understand why the notion of dimension  of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).
Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of the $K$-vector $L$ space depends on possible enlargements of $K$ ?
Finally:
Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

REPLY [3 votes]: If I understand correctly what Question 1 is asking, then there are easy counterexamples even using commutative fields.
Let $K=\mathbb{R}$ and $L=\mathbb{C}$. Then $\begin{pmatrix}1&i\\1&i\end{pmatrix}$ is a matrix over $L$ that has no nonzero elements of $K^2$ in its right kernel, but the element $\begin{pmatrix}1&-1\end{pmatrix}$ of $K^2$ is in its left kernel.