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
  • 1
    $\begingroup$ 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

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.