# Determinant of block matrices.

$$X = \begin{pmatrix} 1+b_1 & 1 & 0 & 0 & 0 & \frac{1}{a_{6}} \\ 1+b_2 & 1 & 1 & 0 & 0 & -\frac{a_1}{a_6} \\ b_3 & 1 & 1 & 1 & 0 & -\frac{a_2}{a_6} \\ b_4 & 0 & 1 & 1 & 1 & -\frac{a_3}{a_6} \\ b_5 & 0 & 0 & 1 & 1 & 1-\frac{a_4}{a_6} \\ b_6 & 0 & 0 & 0 & 1 & 1-\frac{a_5}{a_6} \end{pmatrix}$$

The Schur complement w.r.t. the first and last row/column gives

$$S = \begin{pmatrix} 1+b_1 & \frac{1}{a_6}\\ b_6 & 1 - \frac{a_5}{a_6} \end{pmatrix} -\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 &1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}^{-1} \begin{pmatrix}1+b_2 & -\frac{a_1}{a_6} \\ b_3 & -\frac{a_2}{a_6}\\ b_4 & -\frac{a_3}{a_6}\\ b_5 & 1-\frac{a_4}{a_6}\end{pmatrix}.$$

$$S = \begin{pmatrix} b_1 - b_2 + b_4 - b_5 & - \frac{-1-a_1+a_3-a_4+a_6}{a_6}\\-1-b_2+b_3-b_5+b_6 & \frac{a_1-a_2 + a_4 - a_5}{a_6}\end{pmatrix}$$

Then $$\det(X) = \det\Biggr(\begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}\Biggl). \det(S)$$.

How matrix $$S$$ is obtained? I am not sure why and how to take the blocks here. How $$S$$ is derived? I can see that in matrix $$M$$ but how to approach it in matrix $$X$$?

Suppose $$M = \begin{pmatrix} A & B \\ C & D\end{pmatrix}$$. The Schur complement of $$D$$ w.r.t $$M$$ is given by $$M/D = A - B D^{-1} C$$. It is easy to see when there are contiguous blocks. But when they are not contiguous, how we apply the formula? Like how we get matrix $$S$$. But I am not sure how to connect this with matrix $$X$$.

How matrix $$X$$ will be different from $$Y$$ below, by clubbing the blocks together the blocks used to construct matrix $$S$$.

$$Y = \begin{pmatrix} 1+b_1 & \frac{1}{a_6} & 1 & 0 & 0 & 0 \\ b_6 & 1-\frac{a_5}{a_6} & 0 & 0 & 0 & 1 \\ 1+b_2 & -\frac{a_1}{a_6} & 1 & 1 & 0 & 0 \\ b_3 & -\frac{a_2}{a_6} & 1 & 1 & 1 & 0 \\ b_4 & -\frac{a_3}{a_6} & 0 & 1 & 1 & 1\\ b_6 & 1-\frac{a_4}{a_6} & 0 & 0 & 1 & 1 \end{pmatrix}$$

• What exactly is the question? Your matrix is finite, so you can just use a computer-algebra system like sympy to compute the determinant. If you insist on doing it by hand, notice that the middle part is a tridiagonal matrix which has simple formulas for the determinant and inverse. Aug 31, 2022 at 8:59
• Yes, my question is about how to obtain $S$? which block matrices are considered? To me it seems like six different block matrices are considered for matrix $A$. Aug 31, 2022 at 9:07
• but I ma aware of the usage of 4 block matrices like for matrix $M$ Aug 31, 2022 at 9:19

The blocks in the Schur complement need not be contiguous. In your case, the diagonal blocks corresponds to the submatrices

\begin{aligned} A &=\begin{bmatrix}X_{11} & X_{1n} \\ X_{n1} & X_{nn}\end{bmatrix} =\begin{bmatrix}1+b_1 & \frac{1}{a_6} \\ b_6& 1 - \frac{a_5}{a_6}\end{bmatrix} \\ D&=\begin{bmatrix}X_{2,2} & ⋯ & X_{2,n-1} \\ ⋮&&⋮\\ X_{n-1,2} & ⋯& X_{n-1,n-1}\end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{bmatrix} \end{aligned}

This is justified as follows: Consider the matrix $$M' = P^⊤MP$$ obtained by permuting both the rows and columns of a $$n×n$$ matrix $$M$$ by some permutation matrix $$P∈ℙ_n$$. We can compute the Schur complement of a contiguous block partition of $$M'$$:

\begin{aligned} M' &=\begin{bmatrix}A&B\\C&D\end{bmatrix} =\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix} \begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix} \begin{bmatrix}I_{p}&0\\D^{-1}C&I_{q}\end{bmatrix}. \end{aligned}

But since $$\det(M') = \det(P^⊤ M P) = \det(M)$$ we have

$$\det(M) = \det(D) ⋅ \det(\underbrace{A-BD^{-1}C}_{=M'/D =S})$$

In your case, $$P$$ is the permutation matrix corresponding to the cycle $$π=(2…n)$$. And the size of the diagonal blocks is $$2×2$$ and $$(n-2)×(n-2)$$ respectively.

• Ok, so $S$ is the Schur complement of the tridiagonal matrix with respect to matrix $A$? I am thinking this because inverse of the tridiagonal matrix is taken ? is this correct? Aug 31, 2022 at 9:44
• @BAYMAX $S$ is just the schur complement $M'/D=A-BD^{-1}C$. Aug 31, 2022 at 12:14
• @BAYMAX I took the freedom to relabel some of the variables in both your post and my answer. This was necessary since quantities like "A" occurred multiple times with different meaning. Sep 1, 2022 at 7:42
• @BAYMAX In your problem $A$ and $D$ are the sub-matrices given in the beginning of my post. Again, the point is the determinant of $X$ is the same as the determinant of $X'=P^⊤ X P$ and we perform the contiguous block decomposition on $X'$. You can then get the non-contiguous blocks for $X$ by applying the permutation in reverse. Sep 1, 2022 at 7:46
• @BAYMAX It's really annoying to type large matrices. You just take $X$, swap the the rows $2, ..., n$ in a circular fashion, then do the same for the columns. The resulting matrix is $X'$, and the main blocks are of size $2×2$ and $n-2×n-2$. Sep 1, 2022 at 7:58

I think you have a partition as follows

$$X=\left(\begin{array}{c|cccc|c} 1+b_{1} & 1 & 0 & 0 & 0 & \frac{1}{a_{6}}\\ \hline 1+b_{2} & 1 & 1 & 0 & 0 & -\frac{a_{1}}{a_{6}}\\ b_{3} & 1 & 1 & 1 & 0 & -\frac{a_{2}}{a_{6}}\\ b_{4} & 0 & 1 & 1 & 1 & -\frac{a_{3}}{a_{6}}\\ b_{5} & 0 & 0 & 1 & 1 & 1-\frac{a_{4}}{a_{6}}\\ \hline b_{6} & 0 & 0 & 0 & 1 & 1-\frac{a_{5}}{a_{6}} \end{array}\right)$$

Consider these two projection matrics, $$P$$ a 6×2 matrix, and its orthogonal $$Q$$ a 6×4 matrix.

\begin{aligned}P & =\left(\begin{array}{cc} 1\\ \hline \\ \\ \\ \\ \hline & 1 \end{array}\right) & Q & =\left(\begin{array}{cccc} \\ \hline 1\\ & 1\\ & & 1\\ & & & 1\\ \hline \\ \end{array}\right)\end{aligned}

then you can extract the submatrices as follows

$$\begin{array}{cc} A=P^{\top}XP & B=P^{\top}XQ\\ C=Q^{\top}XP & D=Q^{\top}XQ \end{array}$$

The result is

$$\begin{array}{cc} A=\begin{bmatrix}1+b_{1} & \frac{1}{a_{6}}\\ b_{6} & 1-\frac{a_{5}}{a_{6}} \end{bmatrix} & B=\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\\ C=\begin{bmatrix}1+b_{2} & -\frac{a_{1}}{a_{6}}\\ b_{3} & -\frac{a_{2}}{a_{6}}\\ b_{4} & -\frac{a_{3}}{a_{6}}\\ b_{5} & 1-\frac{a_{4}}{a_{6}} \end{bmatrix} & D=\begin{bmatrix}1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1 \end{bmatrix} \end{array}$$

and the composition of $$X$$ from the submatrices is

$$X=PAP^{\top}+PBQ^{\top}+QCP^{\top}+QDQ^{\top}$$

Considering the above decomposition, the Shur complement is

\begin{aligned}S & =A-B\,D^{-1}C\\ & =\left(P^{\top}XP\right)-\left(P^{\top}XQ\right)\left(Q^{\top}XQ\right)^{-1}\left(Q^{\top}XP\right)\\ & =P^{\top}\left(X-XQ\left(Q^{\top}XQ\right)^{-1}Q^{\top}X\right)P \end{aligned}

Now for $$Y$$ the partition is contiguous I think

$$Y=\left(\begin{array}{cc|cccc} 1+b_{1} & \frac{1}{a_{6}} & 1 & 0 & 0 & 0\\ b_{6} & 1-\frac{a_{5}}{a_{6}} & 0 & 0 & 0 & 1\\ \hline 1+b_{2} & -\frac{a_{1}}{a_{6}} & 1 & 1 & 0 & 0\\ b_{3} & -\frac{a_{2}}{a_{6}} & 1 & 1 & 1 & 0\\ b_{4} & -\frac{a_{3}}{a_{6}} & 0 & 1 & 1 & 1\\ b_{6} & 1-\frac{a_{4}}{a_{6}} & 0 & 0 & 1 & 1 \end{array}\right)$$

which makes the extraction of the sub-matrices trivial.

• What does P{6,2} mean? I could not find the notation here: mathworld.wolfram.com/ProjectionMatrix.html Sep 2, 2022 at 4:54
• I was just noting the size of the $P$ and $Q$ matrices and forgot to update when I edited the post. Sep 2, 2022 at 12:10