3
$\begingroup$

I expect this to be true and proven, but I can't find any proofs of this. So anyone can confirm or deny this?

Let $R$ be a commutative ring, and let $M$ be a $kn\times kn$ matrix, which can be split into $n^{2}$ block of dimension $k\times k$. Assuming the block matrices form a solvable Lie algebra. Then the determinant of $M$ can be found by treating each block as a single element of the matrix ring and use the usual formula for the determinant on that to get a matrix, then calculate the determinant of that matrix.

Thank you.

$\endgroup$
  • $\begingroup$ The counterexample below still seems to work. (Also, there are issues with the non-commutativity of matrix multiplication.) $\endgroup$ – Christian Remling Mar 2 '17 at 1:59
  • $\begingroup$ @ChristianRemling:no it does not work anymore, since after the calculation you get the determinant of diag(1,-1) which is correct. Yes there are non-commutativity issue when you attempt to use the determinant formula, but it should not matter which one you used. $\endgroup$ – user105303 Mar 2 '17 at 2:33
  • $\begingroup$ Try the matrix where the $1$'s are in the $(1,1), (4,4), (2,3), (3,2)$ slots. The formula still fails. $\endgroup$ – Christian Remling Mar 2 '17 at 16:21
  • $\begingroup$ @ChristianRemling: you're right that the example work...in characteristic $2$. Otherwise the blocks don't form a solvable Lie algebra. $\endgroup$ – user105303 Mar 2 '17 at 16:48
1
$\begingroup$

It is true when $R$ is reduced, without $\mathbb{Z}$-torsion. If your blocks are $(M_{i,j})_{1 \leq i,j \leq n}$ and if $$N = \sum_{\sigma \in \mathfrak{S}_n} \epsilon(\sigma) M_{1,\sigma(1)} \dots M_{n,\sigma(n)},$$ then $\mathrm{det}(M) = \mathrm{det}(N)$.

Indeed, if $R$ is reduced without $\mathbb{Z}$-torsion then it can be embedded into a product of algebraically closed fields of characteristic $0$ (take algebraic closures of the residue fields at the minimal primes), so that one can assume that $R$ is an algebraically closed field of characteristic $0$. Then by Lie's theorem on can assume that the blocks $(M_{i,j})$ are all upper-triangular. By replacing $R$ with an algebraic closure of $R(T)$, and $M_{1,1}$ by $M_{1,1} + T I_k$ if necessary, one can assume that $M_{1,1}$ is invertible. The proof is by induction on $n$, but I will detail only the case $n=2$. In this case one has $$ \begin{pmatrix} M_{1,1} & M_{1,2} \\ M_{2,1} & M_{2,2} \end{pmatrix} \begin{pmatrix} I_k & - M_{1,1}^{-1} M_{1,2} \\ 0 & I_k \end{pmatrix} = \begin{pmatrix} M_{1,1} & 0 \\ M_{2,1} & M_{2,2} - M_{2,1} M_{1,1}^{-1} M_{1,2} \end{pmatrix}, $$ so that $\mathrm{det}(M) = \mathrm{det}(M_{1,1}) \mathrm{det}(M_{2,2} - M_{2,1} M_{1,1}^{-1} M_{1,2}) $. Since the blocks are upper triangular, the determinants are computed by taking the product of diagonal entries, so that this is equal to $\mathrm{det}(M_{1,1}M_{2,2} - M_{2,1} M_{1,2})$.

The hypothesis "without $\mathbb{Z}$-torsion" is necessary. Indeed, consider $R = F(T)$ where $F$ is a finite field and consider the $R$-vector space $V = R^F = \mathrm{Maps}(F,R)$. Consider the endomorphisms given by $$ a : f \in V \mapsto (x \mapsto f(x+1)) \\ b : f \in V \mapsto (x \mapsto xf(x)) . $$ Then $[a,b] = a, [a^{-1},b] = - a^{-1}$, so that the Lie algebra generated by $a,a^{-1},b$ is solvable. Consider the block matrix $$ \begin{pmatrix} a & Tb \\ b & a^{-1} \end{pmatrix}. $$ By the computation above, its determinant is $$\mathrm{det}(\mathrm{id} - aba^{-1}(Tb)) = \mathrm{det}(\mathrm{id} - (b+1)bT)) = \prod_{x \in F} (1 - x(x+1) T).$$ This is a polynomial of degree $|F| - 2$. However, $$ \mathrm{det}(a a^{-1} - b (Tb)) = \prod_{x \in F} (1 - x^2 T) $$ has degree $|F| - 1$, hence is not equal to the determinant of our block matrix.

$\endgroup$
  • $\begingroup$ Thank you, the proof looks correct. Do you have any counterexample for nonreduced case? $\endgroup$ – user105303 Mar 2 '17 at 16:49
  • $\begingroup$ @ChristianRemling: the proof here assume characteristic $0$, so your counterexample does not work (and the one in the other answer does not work either way). $\endgroup$ – user105303 Mar 2 '17 at 16:51
  • $\begingroup$ Oh, I got a counterexample by making a small modification to the one above: let $x\not=0$ and $x^{2}=0$ then the matrix with $1$ at $(1,1),(2,3),(4,4)$ and $x$ at $(3,2)$ work. $\endgroup$ – user105303 Mar 2 '17 at 17:26
4
$\begingroup$

It's false.

Consider the permutation matrix associated with the permutation $(24) \in S_4$. The matrix is a 4 by 4 matrix, which can be broken up into 4 2 by 2 blocks. Each of the 2 by 2 blocks is singular (due to having a row consisting entirely of 0s), but the entire matrix has determinant -1.

Edit: apparently I missed the word "solvable" in the question, so this is not an answer.

$\endgroup$
  • $\begingroup$ sorry I mixed up the order, editing it now $\endgroup$ – user105303 Mar 2 '17 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.