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Let $S$ be the subspace of vector space of all real $n\times\ n$ matrices generated by all matrices of the form $AB-BA$ with $A$ and $B$ in vector space all $n\times\ n$. What is the dimension of $S$?

Najib Idrissi
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Pooja
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    Those matrices have zero trace. – Angina Seng Mar 05 '18 at 09:04
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    By the comment above the dimension is at most $n-1$. Can every traceless matrix be written in that form? See for example here https://math.stackexchange.com/questions/729796/traceless-matrices for an answer to that question. – Mathematician 42 Mar 05 '18 at 09:27
  • In fact, you can find this paper online: https://projecteuclid.org/download/pdf_1/euclid.mmj/1028990168 They prove exactly what is needed. The trick is that if $M=AB-BA$ then $P^{-1}MP$ is also a commutator, hence we may assume that $M$ is in rational canonical form. – Mathematician 42 Mar 05 '18 at 09:30
  • The above statement is a bit too strong though as we only need that $\text{Tr}(M)=0\Rightarrow \exists n\in\mathbb{N}: \exists A_i,B_i$ for $i\leq n$ such that $M=\sum_i A_iB_i-B_iA_i$. This statement is proven in this beautiful answer by Mariano here : https://math.stackexchange.com/questions/125219/traceless-matrices-and-commutators It does require some more abstract knowledge though. – Mathematician 42 Mar 05 '18 at 09:44
  • @mathcounterexamples.net Pay attention when you make an edit... You removed the question! – Najib Idrissi Mar 05 '18 at 15:57
  • @NajibIdrissi Oups... In fact I changed the title in order to make it more meaningful. I wasn't aware that this removed the question! – mathcounterexamples.net Mar 05 '18 at 15:59

2 Answers2

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As the matrices are highly non-commutative, the following isomorphism shouldn't be to surprising:

$$\frac{M_n(k)}{[M_n(k),M_n(k)]}\cong k.$$

(This follows as the zeroth cohomology of a $k$-algebra is a Morita invariant). Here $k$ denotes an arbitrary field. Notice that $\text{Tr}:M_n(k)\rightarrow k$ is a surjective linear map. By the first isomorphism theorem, we have that $\frac{M_n(k)}{\ker(\text{Tr})}\cong k$. Since $[M_n(k),M_n(k)]\subseteq \ker(\text{Tr})$ and by the above isomorphism, we conclude that $[M_n(k),M_n(k)]=\ker(\text{Tr})$, which we needed to show.

Mathematician 42
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    don't you need to show, for the first statement, that for any $A$ there are $[B,C]$ such that $A+[B,C]$ is diagonal? How do you prove this directly? – glS Apr 11 '21 at 07:36
  • @glS : Can you be a bit more precise? Which first statement? The first isomorphism $\frac{M_n(k)}{[M_n(k),M_n(k)]}\cong k$? – Mathematician 42 Apr 11 '21 at 08:09
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    yes, that one . – glS Apr 11 '21 at 08:11
  • @glS : Good question. The indirect way is by using that the zeroth Hochschild cohomology is a Morita invariant and that $\text{HH}^0(A)\cong A/[A,A]$ for any $k$-algebra. I'm thinking about a more direct argument without having to compute too much stuf. – Mathematician 42 Apr 11 '21 at 08:22
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    I saw that argument in a linked post, yes. It is however, how shall I put it, quite an unintelligible one for someone (like me) that does not know cohomology theory =) – glS Apr 11 '21 at 08:26
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    @glS : I must admit that a direct proof is not easy, so I lied about being able to this directly. I still like the indirect argument, but it hides a lot of work in it's proofs.

    As far as the question itself goes, the answer given in https://math.stackexchange.com/questions/1007473/determining-the-dimension-of-span-ab-ba-a-b-in-m-n-times-n-mathbb-r seems to be the most direct approach.

    Alternatively, it's clear that $[M_n(k),M_n(k)]\subseteq \ker(\text{Tr})$. One can show the reverse inclusion as in https://people.eecs.berkeley.edu/~wkahan/MathH110/trace0.pdf but it's not straightforward.

     – Mathematician 42 Apr 11 '21 at 08:33
  • @ glS : So concretely, it seems to be easier to first show that $[M_n(k),M_n(k)]=\ker(\text{Tr})$ and then conclude the first isomorphism. Showing this isomorphism directly seems to be difficult and indeed requires work! – Mathematician 42 Apr 11 '21 at 08:42
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Name $E_{ij}=(e_{kl})$ the matrix with all coefficients vanishing except $e_{ij}=1$.

then for $i,j,k,l \in \{1, \dots, n\}$ you have: $$E_{ij}E_{kl} = \delta_{jk}E_{il}$$ where $\delta_{ij}$ is kronecker delta symbol.

Therefore $$E_{ij}E_{kl}-E_{kl}E_{ij} = \delta_{jk}E_{il}-\delta_{li}E_{kj}.$$

Taking $j = k$ and $l \neq i$ you get $$E_{ij}E_{kl}-E_{kl}E_{ij} = E_{il}.$$

Which means that you can get all matrices of the form $E_{il}$ with $l \neq i$ by the operation $AB-BA$. By linear combinations, you get all the matrices having zeros on the diagonal.

Now taking $j = k$ and $l = i$ you get $$E_{ij}E_{kl}-E_{kl}E_{ij} = E_{ii} - E_{jj}.$$ Which means all the matrices with all coefficients equal to zero except a one and a minus one on the diagonal. And those matrices generate by linear combinations all the diagonal matrices with a zero trace.

Combining all the avove, we've proven that the subspace of the matrices space you're looking for is the subspace of matrices with zero trace.

mathcounterexamples.net
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  • is there a typo with the indices in the expression for $E_{il}$? Also, how does the extension by linearity work? If you know that e.g. $E_{12}=[A,B]$ and $E_{21}=[C,D]$, how does that tell you that $E_{12}+E_{21}=[E,F]$ for some $E,F$? – glS Apr 11 '21 at 07:06