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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

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

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