# Determinant of block matrix with singular blocks on the diagonal

Let $A$ and $D$ be square matrices, and let $B$ and $C$ be matrices of valid shapes to allow the formation of $$M = \begin{bmatrix} A & B \\ C & D \end{bmatrix}.$$ If $\det{A}\neq0$, we may use the Schur complement to express $\det{M}$ in terms of its constituent blocks as $$\det{M} = \det{A}\cdot\det(D-CA^{-1}B),$$ and if $\det{D}\neq0$ we have in a similar fashion that $$\det{M} = \det(A-BD^{-1}C)\cdot\det{D}.$$

My question: Does there exist a similar formula expressing $\det{M}$ in terms of its constituent blocks, that is valid in case $\det{A}=\det{D}=0$?

Without loss of generality, let $$A$$ be $$m$$-by-$$m$$ and $$D$$ be $$n$$-by-$$n$$, with $$m\ge n$$. Consider $$f(t)=\det\left( \begin{array}{cc} A+tI_m&B\\ C&D \end{array} \right).$$ Obviously,

• $$f(t)$$ is a polynomial of $$t$$, for which it is analytic on $$\mathbb{R}$$;
• $$f(0)$$ returns the desired determinant;
• $$g(t)=\det\left(A+tI_m\right)$$ is also a polynomial of $$t$$, for which it has isolated zeros;
• $$g(0)=0$$ provided that $$A$$ is singular, yet since $$t=0$$ is an isolated zero of $$g(t)$$, $$g(t)\ne 0$$ for all $$t\in\left(-\delta,\delta\right)\setminus\left\{0\right\}$$ for some $$\delta>0$$.

Now, since $$g(t)\ne 0$$ on some $$\left(-\delta,\delta\right)\setminus\left\{0\right\}$$, it follows that $$A+tI_m$$ is invertible on this domain. Therefore, $$f(t)=\det\left(A+tI_m\right)\det\left(D-C\left(A+tI_m\right)^{-1}B\right).$$ Consequently, the continuity of $$f(t)$$ yields $$f(0)=\lim_{t\to 0}\left(\det\left(A+tI_m\right)\det\left(D-C\left(A+tI_m\right)^{-1}B\right)\right).$$

Now, let us focus on two distinct cases. First, if $$A=O_m$$, i.e., $$A$$ is a zero matrix of order $$m$$. In this case, the above formula gives \begin{align} f(0)&=\lim_{t\to 0}\left(\det\left(tI_m\right)\det\left(D-C\left(tI_m\right)^{-1}B\right)\right)\\ &=\lim_{t\to 0}\left(t^m\det\left(D-\frac{1}{t}CB\right)\right)\\ &=\lim_{t\to 0}\left(t^m\det\left(\frac{1}{t}\left(tD-CB\right)\right)\right)\\ &=\lim_{t\to 0}\left(t^{m-n}\det\left(tD-CB\right)\right). \end{align} Recall that $$m\ge n$$. We therefore obtain

• If $$m>n$$, it is obvious that $$f(0)=0$$;
• If $$m=n$$, it follows that $$f(0)=\det\left(-CB\right)=\left(-1\right)^n\det\left(CB\right)$$.

Second, consider $$A\ne O_m$$. This case is more complicated, and there is no elegant form for $$f(0)$$ with only $$A$$, $$B$$, $$C$$, and $$D$$ involved. However, since $$A\ne O_m$$, we may perform some elementary operations of the second type, i.e., row- and column-switching transformations, such that after the operations, we obtain $$f(0)=\det\left( \begin{array}{cc} A'&B'\\ C'&D' \end{array} \right),$$ where, e.g., $$A'$$ results from $$A$$ by switching $$A$$'s rows and columns, such that $$A''$$, the $$k$$-by-$$k$$ square matrix made up of the first $$k$$ rows and columns of $$A'$$, is invertible. The existence of such an $$A''$$ is guaranteed by the fact that $$A\ne O_m$$. In this way, $$f(0)=\det\left( \begin{array}{cc} A''&B''\\ C''&D'' \end{array} \right).$$ Thanks to the invertibility of $$A''$$, $$f(0)=\det\left(A''\right)\det\left(D''-C''\left(A''\right)^{-1}B''\right).$$ This result is much less elegant. $$A''$$ is only part of $$A$$. $$B''$$ contains part of both $$A$$ and $$B$$, and so does $$C''$$. $$D''$$ contains the whole $$D$$, and part of $$A$$, $$B$$, and $$C$$. Besides, there are switches of rows and columns as well.

• While this answer did not provide a formula for the general case, it did provide a very nice idea which might be possible to use in some important special cases. Congratulations for the first (smaller) bonus! Feb 4, 2019 at 22:50
• @MårtenW: Wow! Thank you for your acknowledgement and your bonus! It is so beyond the generous of you! I hope this partial answer would be somewhat helpful for you. By the way, I think it quite promising that its results for the special cases would help clarify the sign ambiguity in polfosol's answer. Thanks :-) Feb 5, 2019 at 0:49

For $$M=\begin{bmatrix}A & B\\C & D\end{bmatrix}$$ let $$N=M^TM$$. So that $$N:=\begin{bmatrix}E & F\\F^T & G\end{bmatrix}$$ where \begin{align} E&=A^T A+C^T C\\F&=A^T B+C^T D\\G&=B^T B+D^T D \end{align} Now, if $$M$$ is non-singular then $$N$$ is a positive definite matrix. Hence according to this other post, $$E$$ is a positive definite (i.e. non-singular) block matrix.

Since $$E$$ is non-singular, we can use Schur complement to obtain $$\det N$$. $$\det{N} = \det{E}\cdot\det(G-F^T E^{-1}F)=(\det M)^2$$ The only remaining part in this case is to determine the sign of $$\det M$$, which apparently, there is no easy way to do that in general.

We might be able to get the sign of $$\det M$$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.

• Nice answer (+1) and nicer nickname by the way :) Feb 1, 2019 at 21:44
• Except for leaving that sign ambiguity, this answer does precisely what I was asking for, and for that reason I'm going to award the bigger bounty as soon as the system lets me. Feb 4, 2019 at 22:47

Hint: $$M = \begin{bmatrix} A & B \\ C & D \end{bmatrix}. = \begin{bmatrix} 0 & I \\ I & 0 \end{bmatrix}. \begin{bmatrix} C & D \\ A & B \end{bmatrix}$$

Where $$I$$ is the idenity matrix of appropriate size.

• I thought about the same hint, but C may not even be square, so this is a dead end in general.
– Dirk
Jan 29, 2019 at 6:33