Determinant of block matrices.

Question

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

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 $S$.

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

Answer 1

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

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

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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.

Answer 2

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.