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The claim is made here that for $k$ a field

The left ideals of $M_n(k)$ are all of the form $$\{A \in M_n(k) \mid \operatorname{ker}A > S \}, \rlap{ \qquad \text{for some subspace $S$.}}$$

I was trying to think through that claim. I understand why all the left ideals of $M_n(k)$ are of the form $$\{A \in M_n(k) \mid \operatorname{Rowspace}(A) < S \}, \rlap{ \qquad \text{for some subspace $S$.}}$$

In the presence of an inner product, we have that $\operatorname{ker}A^T = \operatorname{Range}(A)^{\bot}$, and therefore we get the characterization we want. But what about if there isn't an inner product?

Does the same hold for $M_n(\Delta)$, where $\Delta$ is a division ring?

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Eric Auld
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  • It should be the same via some argument not relying on inner products. I'll definitely let you know if I come across a source – rschwieb Jan 21 '17 at 23:14
  • Do you really need an inner product $(u,v)\mapsto v^Tu$? Isn't it enough to consider only linear functionals of the form $L_v:,u\mapsto v^Tu$? You still have $\ker A^T=\operatorname{ann}\left(\operatorname{range}(A)\right)$. – user1551 Jan 22 '17 at 02:35
  • Maybe this one is useful: http://math.stackexchange.com/questions/1586280/any-left-ideal-of-m-n-mathbbf-is-principal – Sungjin Kim Feb 12 '17 at 01:31
  • It all depends on the ambiguous definition of the kernel of a matrix, it considered as working on row vectors then the statement using the Kernel is absulutely correct : every row of any element $M$ of the ideal has rows that are combination of fixed rows $r_1, r_2, \ldots$ so for some combination $(a_1r_+a_2r_2+\ldots)M = 0$. – Marc Bogaerts Feb 18 '17 at 21:29
  • @i707107 That's awesome. – Eric Auld Feb 18 '17 at 21:34

1 Answers1

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Using the theory of nondegenerate symmetric bilinear forms, it is possible to show that $\operatorname{ker}A^T = \operatorname{Range}(A)^{\bot}$ without the need for existence of an inner product (when we are discussing matrices over a field). The symbol $\bot$ would then represent an orthogonal complement with respect to the relevant nondegenerate symmetric bilinear form. Here is a copy from some work of mine showing that the row space of a matrix is the right orthogonal complement of the null space with respect to what I would regard as the most natural nondegenerate symmetric bilinear form:

Consider the function $f: \mathscr{F}^n \times \mathscr{F}^n > \rightarrow \mathscr{F}$, defined by $f(v,w)=v^Tw$. It is easy to see that:

  1. $f_{w_0}:\mathscr{F}^n \rightarrow \mathscr{F}$ defined by $f_{w_0}(v)=f(v,w_0)$ for any given $w_0 \in \mathscr{F}^n$ is a linear functional and therefore belongs to the dual space of $\mathscr{F}^n$.
  2. Similarly $f_{v_0}:\mathscr{F}^n \rightarrow \mathscr{F}$ defined by $f_{v_0}(w)=f(v_0,w)$ for any given $v_0 \in \mathscr{F}^n$ is a linear functional and therefore belongs to the dual space of $\mathscr{F}^n$.

By the properties above, $f$ is a bilinear form, and furthermore it is a symmetric bilinear form since $f(v,w)=f(w,v)$ for any $v,w \in \mathscr{F}^n$. A symmetric bilinear form $g:\mathcal{V} \times \mathcal{V} \rightarrow \mathscr{F}$ is non-degenerate if and only if, for every nonzero $v \in \mathcal{V}$ there exists $w \in \mathcal{V}$ so that $g(v,w) \neq 0$ [p.455]{golan}. Let $v=(v_1,v_2,\ldots,v_n)^T \in \mathscr{F}^n$ be nonzero. Suppose $i \in \{1,2,\ldots,n\}$ is the least integer so that $v_i$ is nonzero. Let $w=(0,\ldots,0,v_i,0,\ldots,0)^T \in \mathscr{F}^n$ (with $v_i$ in entry $i$ of $w$). Then $$f(v,w)=v^Tw=v_i^2 \neq 0,$$ since a field contains no divisors of zero. This proves that $f$ is non-degenerate for any field $\mathscr{F}$.

The right $f$-orthogonal complement [p.458]{golan} of row$(G)$ is \begin{eqnarray} \nonumber \text{row}(G)^{\perp_f}&=&\{w \in \mathscr{F}^n:f(v,w)=0 \text{ for all } v \in \text{row}(G) \} \\ \nonumber &=& \{w \in \mathscr{F}^n:Gw=0 \} \\ &=& \text{N}(G). \end{eqnarray}

The theory can be studied in detail on page 455-458 of this textbook, which is the source referenced above. As you can see $f(v,w)$ is nondegenerate due to the fact that a field has no divisors of zero.

For a division ring the symmetric property would fail in general, but I don't actually think this is absolutely necessary: You can see in the proof above I mention the right $f$-orthogonal complement, and the last part would still hold in the absence of commutativity.

Christiaan Hattingh
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