Elementary linear algebra over a (possibly skew) field $K$

Question

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 $K$ ?

Answer 1

Re: question 2, the rank of a free module over any ring $R$ with IBN is well-defined, and invariant under arbitrary extensions of scalars $f : R \to S$ where $S$ also has IBN, because of the straightforward isomorphism

$$R^n \otimes_R S \cong S^n.$$

In particular, division rings have IBN, as do commutative rings.

A different and inequivalent question is why the dimension of the kernel of a linear transformation is unchanged by extension of scalars (for a map $f : R\to S$ of division rings). The reason is that a division ring will always be flat as a module over a division ring.

Re: question 3, it depends on what you mean by "enlarging" a Zariski closed subset of affine space via an extension of scalars $k \to K$. One thing you might mean is taking some finitely generated $k$-algebra and considering first its set of $k$-points and then its set of $K$-points. Then it is not true that the dimension is invariant because e.g. the set of $k$-points could be empty while the set of $K$-points is nonempty (consider the variety $x^2 + y^2 = -1$ over $\mathbb{R}$ and then over $\mathbb{C}$).

Answer 2

For approaches of (2) i.e. resolution of systems with coefficients in a division ring (a skew field), you can consult

[1] A. Heyting, Die Theorie der linearen Gleichungen in einer Zahlen- spezies mit nichtkommutativer Multiplikation, Math. Ann., 98, 465- 490 (1927).

[2] A.R. Richardson Simultaneous linear equations over a division ring. Proc. Lond. Math. Soc., 28, 395-420 (1928).

Now, you have also the non-commutative determinants of Gelfand and Retakh

arXiv:math/0208146

you will find in there a nice historical introduction on the subject of noncommutative determinants as well as a construction of the free division ring.

and in their book. Applied later to the theory of noncommutative symmetric functions started by Gelfand, Krob, Lascoux, Leclerc, Retakh, Thibon (up to my knowledge continued through seven papers).

A warning about the rank Let $\Gamma=F(a,b)$ be the free group on two letters and order it with a total (group) ordering as it can be done this though series, see e.g.

G. Duchamp, J.-Y. Thibon, Simple orderings for free partially commutative groups , International Journal of Algebra and Computation 2 No.3 (1992).

Then consider the skew field $\mathbb{Q}((a,b))$ (Malcev Neumann series, for example as in arXiv:math/0405133) which is the set of functions $\Gamma \rightarrow \mathbb{Q}$ with well-ordered supports (and usual operations). Then, the matrix $$ \begin{pmatrix} ba & a\\ b^2 & b \end{pmatrix} $$ has its columns left proportional but not right proportional. So, the vector space generated by the columns on the left has dimension one and on the right has dimension 2. In fact $M$ is (two-sided) invertible. One has $$ M^{-1}= \begin{pmatrix} [b,a]^{-1} & [a,b]^{-1}ab^{-1}\\ -b[b,a]^{-1} & -b[a,b]^{-1}ab^{-1}+b^{-1} \end{pmatrix} $$ and $M$ is a (two-sided) a zero divisor for the opposite field $K^{op}$ Concerning question 1 using elementary operations, one can prove

Proposition: Let $K$ be a (skew) field and $M\in K^{n\times n}$ (square matrix of dimension $n$). The following are equivalent

1. $M$ is right invertible
2. $M$ is left invertible
3. $M$ is not a right zero divisor
4. $M$ is not a left zero divisor
5. For $\lambda\in K^{1\times n}$ (a row) one has $$ \lambda M=0\Longrightarrow \lambda=0 $$
6. For $\gamma\in K^{n\times 1}$ (a column) one has $$ M\gamma=0\Longrightarrow \gamma=0 $$

As regards question 2, in the general case the adapted notion is that of non-commutative (or quasi-)determinants of Gelfand and Retakh (see above) for example $M$ has four quasi-determinants given by its inverse.

A bit on the relation between right-left kernels-images I pursue a bit for the sake of completeness.

As remarked by Sebastian Goette and ACL, a matrix $M\in K^{n\times n}$ acts on the left on the space of columns $K^{n\times 1}$ (considered as a right $K$-vector space), defining then an element of $\mathrm{End}_K(K^{n\times 1})$, this correspondence is an isomorphism and allows to speak of $ker(M)$ and $Im(M)$ which will be denoted $rker(M)$ and $rIm(M)$ (to express that it is devoted to the right structures).

Likewise, $M$ acts on the right on the space of rows $K^{1\times n}$ (considered as a left $K$-vector space), and the correspondence $\mathrm{End}_K(K^{1\times n})$, is also an isomorphism. Hence the notations $lker(M)$ and $lIm(M)$ (for the same reason).

Now, you have the non-degenerate pairing (still by matrix multiplication) $$ \langle\ |\ \rangle\ :\ K^{1\times n}\otimes_K K^{n\times 1}\rightarrow K^{1\times 1}\simeq K $$ (this time, the two spaces are considered as $K-K$-bimodules).

One can check easily that $lker(M)=(rIm(M))^\perp$ and $rker(M)=(lIm(M))^\perp$. This, with the classic $$ dim(xker(M))+dim(xIm(M))=n $$
where $x$ is one of the symbols $\{l,r\}$ allows to see geometrically that $dim(lker(M))=dim(rker(M))$ and
$dim(lIm(M))=dim(rIm(M))$, this last quantity should be considered as the rank of the matrix $M$.

Answer 3

Regarding Question 2, there is the Dieudonné determinant.

Answer 4

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.

Answer 5

The trick for question (1) is not to use determinants. To select a basis of a vector field, you do not need commutativity. As a consequence, all well-known formulas like $$\dim (V/U)=\dim V-\dim U$$ (if $V$ is finite-dimensional and $U\subset V$ a linear subspace) still hold. For linear maps $F\colon V\to W$, you still have $\mathrm{im} F\cong V/\ker F$, so $\mathrm{rk} F=\dim\mathrm{im} F=\dim V-\dim\ker F$.

More concretely, one can define the rank of an $m\times n$-matrix $A$ over a skew field $K$ as $\mathrm{rk} A=\dim(\mathrm{im} A)$. One can compute it (e.g. using the Gauss algorithm), just as one would over a commutative field. It does not change if one passes from $K$ to $L$. Commutativity is not used in any of the proofs.

So, start with a matrix $A$. Left multiplication by so-called elementary $m\times m$-matrices describes row operations. All elementary matrices $C$ are invertible in the sense that there is another elementary matrix $D$ satisfieing $CD=DC=E_m$ (the only commutativity properties ever used are $k\cdot k^{-1}=k^{-1}k=1$ and $1\cdot k=k\cdot 1=k$, which hold in division rings).

Assume that you arrive at a matrix of the form $$B=CA\begin{pmatrix}\cdots&0&1&*&&\cdots\\&&\cdots&&0&1&*\cdots\\&&&&&\ddots\end{pmatrix}\;,$$ where $C$ is an (invertible) product of elementary matrices. Then it is an exercise to read of $\dim\mathrm{im}B$, $\dim\ker B$ etc., and to check that one gets the same numbers for $A$. Also, the whole computation is valid as well over $L\supset K$.

Edit Alternatively, you can multiply the matrix $B$ above with elementary $n\times n$-matrices from the right (this corresponds to column operatorations) until it becomes a block matrix $$Z=BD=CAD=\begin{pmatrix}E_k&0\\0&0\end{pmatrix}\;,$$ where $D$ is an (invertible) product of elementary matrices and $E_k$ is a unit matrix and $k$ is the rank. The rank is the dimension of the image both on column vectors (where $A$ acts from the left) and on row vectors (where $A$ acts from the right).

For a column $v\in K^n$, you get $Av=0$ if and only if $Z(D^{-1}v)=0$, so $\dim\ker A=\dim\ker Z$. Similarly, $\dim\mathrm{im}A=\dim\mathrm{im}Z$ on columns. If one applies $A$ on rows $\alpha$, then for example $$\beta=\alpha A\Longleftrightarrow \beta=\alpha A=(\alpha C^{-1}) (ZD^{-1})\;,$$ so again $\dim\mathrm{im}A=\dim\mathrm{im}Z$ and $\dim\ker A=\dim\ker Z$ on row vectors.