Given a $4 \times 4$ matrix $S$ over a commutative ring $R$. I want to consider it as a $2\times 2$ matrix over $M_2(R)$. Lets say $S=\left(\begin{array}{cc} A&B \\\ C&D\end{array}\right)$ with entries $A,B,C,D\in M_2(R)$. Asssume further, that they all commute.

So my question is: Can one then compute the determinant stepwise via


I checked this with maple for the case, where all the sub-matrices lie in the commutative subring $\{\left(\begin{array}{c}a&b\\\ 2b&a \end{array}\right)|a,b\in \mathbb{Z}\}$ of $M_2(\mathbb{Z})$. And I also would like to consider the more general case, where one considers a $mn\times mn$ matrix over $R$ as a $m\times m$ matrix over $M_n(R)$.

  • $\begingroup$ OK it also holds, when one replaces $2$ in the upper example by any natural number. $\endgroup$ – HenrikRüping Dec 10 '10 at 15:50
  • $\begingroup$ "Assume further, that they all commute." -- Do you have a class of examples for which this very-special-looking situation occurs? Your question title would seem to be much more general. $\endgroup$ – Allen Knutson Dec 10 '10 at 16:36

The answer is Yes. This is done in Exercise 120 of my web page link text. You can replace $4$ and $2$ by numbers $n$ and $m$ with $m$ dividing $n$.

Later. The required complement. Let me take the situation of the question, but with $A,B,C,D$ being $m\times m$ matrices (thus $n=2m$). If $A$ has an inverse, the Schur complement formula tells you that $$\det S=\det A\cdot\det(D-CA^{-1}B).$$ Because $A$ and $C$ commute, this is nothing but $$\det A\cdot\det(A^{-1}(AD-BC))=\det(AD-BC).$$ Now, what if neither $A$ nor $B,C,D$ admit an inverse ? Just do the following. Replace $\mathcal R$ by $\mathcal R(X)$ (rational fractions). Then apply the formula to $$\Sigma=\begin{pmatrix} A+XI_m & B \\\\ C & D \end{pmatrix},$$ remarking that $A+XI_m$ is non-singular, because its determinant is invertible.

Even later. Here are the details when $p:=n/m$ is larger than $2$. The matrix $S$ is $n\times n$, with commuting blocks $A_{ij}\in M_m(\mathcal R)$, $1\le i,j\le p$. To avoid confusion, I use the notation ${\rm Det}$ to denote the determinant in the commutative subring $\mathcal R'$ of $M_m(\mathcal R)$ spanned by the blocks $A_{ij}$, whereas $\det$ is the determinant of an ordinary matrix.

The proof is an induction over $p$. As in the case $p=2$, I may assume that $A_{11}$ is invertible. Schur's formula gives $$\det S=\det A_{11}\det S_{11},$$ where $$ \qquad S_{11}:=\begin{pmatrix} A_{22} & \cdots & A_{2p} \\\\ \vdots & & \vdots \\\\ A_{p2} & \cdots & A_{pp} \end{pmatrix}-\begin{pmatrix} A_{21} \\\\ \vdots \\\\ A_{p1} \end{pmatrix} A_{11}^{-1}\begin{pmatrix} A_{12} & \cdots & A_{1p} \end{pmatrix}.$$ Because $A_{11}$ commutes to every $A_{ij}$, we find $$\det S ( \det A_{11})^{p-2} = \det T ,$$ where $$ \qquad T= block(A_{11}A_{ij}-A_{i1}A_{1j})_{2\le i,j\le p}.$$ From the induction hypothesis, and because the blocks $A_{11}A_{ij}-A_{i1}A_{1j}$ commute to each other, one has $$\det T=\det {\rm Det}((A_{11}A_{ij}-A_{i1}A_{1j}))_{2\le i,j\le p}.$$ Now, in every commutative ring, we have $$\det((x_{11}x_{ij}-x_{i1}x_{1j}))_{2\le i,j\le p}=x_{11}^{p-2}\det X.$$ Therefore $$\det T=\det\left(A_{11}^{p-2}{\rm Det} S\right)=(\det A_{11})^{p-2}\det{\rm Det} S.$$ Simplifying by $(\det A_{11})^{p-2}$, we obtain the expected formula $$\det S=\det{\rm Det} S.$$

  • $\begingroup$ Could you be more precise about the way you use Schur's (complement?) formula`? $\endgroup$ – darij grinberg Dec 10 '10 at 16:35
  • $\begingroup$ Thanks, but I know of this. The question is how you apply it to the $n>2m$ case. $\endgroup$ – darij grinberg Dec 10 '10 at 20:00
  • $\begingroup$ @darij. See my new edition. $\endgroup$ – Denis Serre Dec 11 '10 at 9:25
  • $\begingroup$ Thanks a lot! I didn't believe the Schur complement works for matrices that are not square... stupid me. I've fixed your text in a few places. $\endgroup$ – darij grinberg Dec 11 '10 at 14:13

You are asking for a polynomial equality in 16 variables. A polynomial equality holds if it holds on an open subset of affine 16-space. So you can restrict to the open subset of 16-space where A, B, C and D have distinct eigenvalues. Now because they commute they can be simultaneously diagonalized (perhaps over a bigger field, but that's okay). The truth of your equation is not affected if we apply this diagonalization (you are conjugating each matrix by a single matrix $X$, which does not change either side of the equation). So you can assume A, B, C and D are all diagonal, which makes this an easy computation.

  • $\begingroup$ I don't believe you need the bigger field for the diagonalization step. $\endgroup$ – Thierry Zell Dec 10 '10 at 18:47
  • $\begingroup$ Huh? He needs it for the eigenvalues to exist. $\endgroup$ – darij grinberg Dec 11 '10 at 14:14
  • $\begingroup$ @Thierry: it may well be the case that none of the matrices is diagonalizable over the initial field. For example, suppose all 4 matrices are rotation matrices in $\mathbb R^2$. $\endgroup$ – Mariano Suárez-Álvarez Dec 11 '10 at 14:15
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    $\begingroup$ By the way, the parameter space for this question is not affine 16-space, since the matrices are required to commute. The variety of tuples of commuting matrices being (generally) complicated (I believe), I am not sure if this argument can be salvaged. (Exercise: count the weasel words in the preceding sentence.) $\endgroup$ – darij grinberg Jun 4 '13 at 17:10
  • $\begingroup$ darij: At the very least, you are right that this needs more argument. I will try to find time to try to think about how to try to supply that argument. $\endgroup$ – Steven Landsburg Jun 4 '13 at 17:15

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