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 $K$ ?


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 $K$ ?
  • $\begingroup$ By left kernel do you mean the space (left-module) of rows $\lambda\in K^{1\times n}$ (one row n columns) such that $\lambda M=(0)$ ? and by right kernel, I think you mean $\gamma\in K^{n\times 1}$ (one column n rows) such that $M\gamma=(0)$ $\endgroup$ Commented Feb 18, 2016 at 17:48
  • $\begingroup$ @QiaochuYuan ok! I meant that the action of $K$ on $\mathcal A$ is two sided. I edited it. $\endgroup$
    – Drike
    Commented Feb 18, 2016 at 18:38
  • $\begingroup$ @DuchampGérardH.E. Yes, precisely ! Except that the coefficients may be in $\mathcal A$. $\endgroup$
    – Drike
    Commented Feb 18, 2016 at 18:44
  • $\begingroup$ There's no particular question to expect that something like question 1 holds. I don't know what you mean by "the action of $K$ is two sided." Do you mean that $A$ is a monoid object in the monoidal category of $(K, K)$-bimodules? $\endgroup$ Commented Feb 18, 2016 at 19:33
  • $\begingroup$ @QiaochuYuan I simplified the question. Now it is about division ring only. $\endgroup$
    – Drike
    Commented Feb 18, 2016 at 20:43

5 Answers 5


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 $\mathbb{C}$).

  • $\begingroup$ So there is a big difference between a subset $A$ of $K^n$ defined as the zero set of an affine map with coefficient in $K$, and one subset $B$ of $K^n$ defined as the zero set of an arbitrary polynomial map with coefficients in $K$. Where as $\dim A$ does not depend on being considered in $L^n$ for some extension field $L/K$, $\dim B$ depends on $K$. $\endgroup$
    – Drike
    Commented Feb 18, 2016 at 20:47
  • $\begingroup$ @Drike Algebraic geometry over skew fields is extremely difficult. Take the solution space of $X^2=-1$ over the quaternions. It is given by the unit sphere $S^2$ inside the imaginary quaternions. So from a real point of view, its quaternionic dimension should be $\frac12$. Another reason is that the algebra of (noncommutative!) polynomials has a rather complicated structure. $\endgroup$ Commented Feb 20, 2016 at 10:59
  • $\begingroup$ @Drike What do you mean when you consider polynomial equations in a non-commutative ring or field? You can of course evaluate a polynomial at a point, but the evaluation map $P\mapsto P(x)$ is not a morphism of rings, so the usual habits do not function well. In you question (3), you had (carefully) restricted yourself to commutative fields. $\endgroup$
    – ACL
    Commented Feb 20, 2016 at 14:01
  • $\begingroup$ @ACL In my comment, related to Question 3, I am considering commutative fields. $\endgroup$
    – Drike
    Commented Feb 20, 2016 at 16:46
  • $\begingroup$ @QiaochuYuan. Dear Yuan, Question 3 still puzzles me : here is how I define the dimension in $K^n$ of an affine algebraic variety $V(S)$ (the zero set in $K^n$ of a finite subset of $S\subset K[x_1,\dots,x_n]$) : the transcendance degree of $K[x_1,\dots,x_n]/\langle S\rangle$ over $K$. With this definition, the variety defined by $x^2+y^2+1=0$ has dimension 1 in both $\mathbf C^2$ and $\mathbf R^2$ doesn't it ? If $K/L$ is a field extension, the transcendance degree of $K[x_1,\dots,x_n]/\langle S\rangle$ over $K$ and the one of $L[x_1,\dots,x_n]/\langle S\rangle$ over $L$ are the same $\endgroup$
    – Drike
    Commented Mar 1, 2016 at 13:43

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


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 $M$.

  • 1
    $\begingroup$ I am confused now. Why would you consider columns as a left vector space at all? I also don't think that the arguments in my answer below fail for your matrix, call it $A$. The image of $A$ acting from the right on row vectors is spanned by the rows viewed as a left vector space, and they are linearly independent, so the row rank is 2. The image of $A$ acting on columns from the left is spanned by the columns, viewed as a right vector space, and has dimension 2 again. If you transpose the matrix, both ranks become 1. So we learn that transposition is dangerous, see @ACL s comment on my answer. $\endgroup$ Commented Feb 20, 2016 at 11:08
  • 1
    $\begingroup$ No, do not be confused because this does not invalidate your arguments and my warning is just to prevent people to treat lightheartedly rank notions. I'll put more hints in my answer. $\endgroup$ Commented Feb 20, 2016 at 12:28
  • $\begingroup$ @SebastianGoette Of course, I had seen the ACL comments, but I wanted to give an explicit example. $\endgroup$ Commented Feb 20, 2016 at 12:54

Regarding Question 2, there is the Dieudonné determinant.

  • $\begingroup$ Is this related to the first algebraic $K$-group of $K$ for invertible matrices? $\endgroup$ Commented Feb 20, 2016 at 11:29

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.


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.

  • 1
    $\begingroup$ When talking about matrices with coefficients in a (possibly non-commutative) ring $A$, it is essential to view $A^n$ as a right $A$-module solely, so that $M_{m,n} (A)$ are exactly $A$-linear maps from $A^n$ to $A^m$. So the images, kernels, etc. you consider are relative to the right $A$-module structure on $A$. If (as Drike asked) one wants to consider left/right kernels, transposition does not help, because it naturally goes from $M_{m,n}(A) $ to $M_{n,m}(A^o)$ (the opposite ring). $\endgroup$
    – ACL
    Commented Feb 20, 2016 at 9:47
  • $\begingroup$ @DuchampGérardH.E. As ACL points out, the column vectors form a right $K$-module. You can get a bimodule by letting $K$ act by mupliples of the identity matrix from the left, but this does not commute with other matrices unless $K$ is commutative. In other word, you have an $M_{n\times n}(K)\times K$-bimodule structure that you can restrict to an $K\times K$-bimodule. $\endgroup$ Commented Feb 20, 2016 at 10:36
  • $\begingroup$ @ACL As you can see from the edit, it is not necessary to transpose any matrix or go to $A^{\mathrm{op}}$. $\endgroup$ Commented Feb 20, 2016 at 10:53
  • $\begingroup$ OK, but what you write (which is true) is (I believe) not what Drike asked. In other words, the right kernel is related to the right image, but may have nothing to do with the left kernel (which, itself, is related to the left image). $\endgroup$
    – ACL
    Commented Feb 20, 2016 at 11:17
  • 1
    $\begingroup$ @SebastianGoette Maybe you're right. But if the $K$ is just a typo then it seems odd to write "there is a non-trivial point of $K^n$ in the left-kernel" instead of "the left-kernel is non-trivial". I hope the OP will clarify. $\endgroup$ Commented Feb 20, 2016 at 12:12

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