why am I getting the correct determinant?

Question

im reading a paper in which the author compute the determinant of this block matrix$$\left( \begin{array}{c|c|c|c|c} X & 0 & 0 & 0 & V_0 \\ \hline -I & X & 0 & 0 & V_1 \\ \hline 0 & -I & X & 0 & V_2 \\ \hline 0 & 0 & -I & X & V_3 \\ \hline 0 & 0 & 0 & -I & X+V_4 \end{array}\right) $$ and he get $\det(V_0+V_1x+V_2x^2+V_3x^3+V_4x^4+x^5)$using the schur compliment which i didn't understand .

so i tried to compute it otherwise and in the process i mistakenly use the determinant expansion by fifth column and get the correct result !!

so i thought the expansion must apply in case of block matrices and i look over the internet and did not find a proof,

sorry $I$ is an identety matrix and $X=x*I$

Answer 1

What you said isn't quite accurate. The determinant of that matrix is not $V_0+xV_1+x^2V_2+x^3V_3+x^4V_4+x^5I$ but $\color{red}{\det}(V_0+xV_1+x^2V_2+x^3V_3+x^4V_4+x^5I)$. In other words, to calculate the determinant of the large matrix, you first treat each sub-block as a scalar and calculate the "determinant" of a $5\times5$ matrix using "blockwise Laplace expansion". The resulting "determinant" is yet another matrix whose size is identical to the size of each sub-block. Take the determinant of this resulting matrix, you get the determinant of the original large matrix.

This blockwise Laplace expansion works because all sub-blocks in the first four block columns commute.

More specifically, consider a block matrix $$ M=\pmatrix{A_{11}&\cdots&A_{1n}\\ \vdots&&\vdots\\ A_{n1}&\cdots&A_{nn}}, $$ where all sub-blocks $A_{ij}$s are $m\times m$ matrices over some commutative ring $K$. If we remove the $i$-th block row and the $j$-th block column from $M$, we get an $m(n-1)\times m(n-1)$ submatrix over $K$, or an $(n-1)\times(n-1)$ matrix over $M_m(K)$ if you view each sub-block of $M$ as a matrix element. Let us call this matrix $\widehat{A}_{ij}$. E.g. $$ \text{if }\ M=\pmatrix{A_{11}&A_{12}&A_{13}\\ A_{21}&A_{22}&A_{23}\\ A_{31}&A_{32}&A_{33}}, \ \text{ then }\ \widehat{A}_{23}=\pmatrix{A_{11}&A_{12}\\ A_{31}&A_{32}}. $$ Now suppose we want to perform a blockwise Laplace expansion of this $M$ along the last column. Two issues immediately arise:

1. How to calculate a blockwise "minor"? Suppose the pivot is $A_{23}$ and we need to calculate the $(2,3)$-th "minor", i.e. we need to calculate the blockwise "determinant" of $\widehat{A}_{23}$, as if each of $A_{11},A_{12},A_{31},A_{32}$ is a scalar. However, the sub-blocks in general do note commute. Shall we calculate the "minor" as $A_{11}A_{32}-A_{12}A_{31},\ A_{11}A_{32}-A_{31}A_{12},\ A_{32}A_{11}-A_{12}A_{31}$ or $A_{32}A_{11}-A_{31}A_{12}$?
2. To differentiate blockwise determinant from the usual determinant, let us denote it as $Det$ (where the letter "D" is capitalised) rather than $\det$. So, unlike $\det(\widehat{A}_{ij})$, which is a scalar in $K$, the blockwise minor $Det(\widehat{A}_{ij})$ is an $m\times m$ matrix over $K$. Again, since this minor does not necessarily commute with its pivot sub-block, it is not immediately clear how the blockwise Laplace expansion should be carried out. Is it \begin{align} &A_{31}Det(\widehat{A}_{31})-A_{32}Det(\widehat{A}_{32})+A_{33}Det(\widehat{A}_{33}),\text{ or}\\ &Det(\widehat{A}_{31})A_{31}-Det(\widehat{A}_{32})A_{32}+Det(\widehat{A}_{33})A_{33}? \end{align}

Fortunately, these issues go away if all sub-blocks outside the row/column that we want to expand along commute.

Note that when all sub-blocks of $\widehat{A}_{ij}$ commute, $Det(\widehat{A}_{ij})$ is well defined. Actually it is just the usual determinant, except that the underlying commutative ring is not $K$ but the ring generated by all sub-blocks of $\widehat{A}_{ij}$. E.g. in our previous example, when $A_{11}A_{32}=A_{32}A_{11}$ and $A_{31}A_{12}= A_{12}A_{31}$, it does not matter whether we calculate $Det(\widehat{A}_{23})$ as $A_{11}A_{32}-A_{12}A_{31}$ or $A_{32}A_{11}-A_{31}A_{12}$, because the two expressions are identical. So, the first issue is resolved.

Concerning the second issue, we define blockwise Laplace expansion as follows:

• Suppose all sub-blocks of $M$, except perhaps those on block column $j$, commute. That is, suppose $A_{rk}A_{sl}=A_{sl}A_{rk}$ whenever $k,l\ne j$. Then the blockwise Laplace expansion of $M$ along block column $j$, denoted by $L_{\ast j}(M)$, is defined as follows: $$ L_{\ast j}(M)=\sum_{i=1}^n (-1)^{i+j}Det(\widehat{A}_{ij})A_{ij}\in M_m(K).\tag{1} $$
• Suppose all sub-blocks of $M$, except perhaps those on block row $i$, commute. That is, suppose $A_{rk}A_{sl}=A_{sl}A_{rk}$ whenever $r,s\ne i$. Then the blockwise Laplace expansion of $M$ along block row $i$, denoted by $L_{i\ast}(M)$, is defined as follows: $$ L_{i\ast}(M)=\sum_{j=1}^n (-1)^{i+j}A_{ij}Det(\widehat{A}_{ij})\in M_m(K).\tag{2} $$ (Note that the order of multiplication of the pivot and the minor in row expansion is opposite to that in column expansion.)

Then we have the following result for column expansion: $$ \det(M) = \det(L_{\ast j}(M))\ (\in K).\tag{$\dagger$} $$ The result for row expansion is similar. (In your particular case, $L_{\ast5}(M)=V_0+xV_1+x^2V_2+x^3V_3+x^4V_4+x^5I_m\in M_m(K)$ and $\det(M)=\det(L_{\ast5}(M))=\det(V_0+xV_1+x^2V_2+x^3V_3+x^4V_4+x^5I_m)\in K$. Since the sub-blocks in the first four columns also commute with those in the fifth one, it doesn't really matter if you put the pivot before or after the minor in $(1)$ here, but in general, you should put the minors before the pivots when you do a column expansion.)

This generalises the theorem (cf. John Silvester, Determinants of Block Matrices) that $\det\pmatrix{A&B\\ C&D}=\det(AD-CB)$ when $AC=CA$.

The proof of $(\dagger)$ is parallel to Silvester's proof. We first redefine $M$ as $M+yI$, where $y$ is an indeterminate adjoined to the ring $K$. By doing so, we can guarantee in our proof that the determinants or blockwise determinants of some proper principal submatrices are not zero divisors. Once the Laplace expansion formula is proved to be true for this new $M$, it is also true for the original $M$ by specialising $y$ to zero.

Now, suppose we want to expand along the last column and all sub-blocks off the last column commute. Denote by $R$ the commutative subring generated by the sub-blocks in the first $n-1$ block columns of $M$. Then $\sum_{i=1}^n(-1)^{i+n}Det(\widehat{A}_{in})A_{ij}=0$ for every $j<n$, because $\left((-1)^{1+n}Det(\widehat{A}_{1n}),\ldots,(-1)^{n+n}Det(\widehat{A}_{nn})\right)$ is just the last row of the adjugate matrix of $$ B=\begin{bmatrix}A_{11}&\cdots&A_{1,n-1}&0\\ \vdots&&\vdots&\vdots\\ A_{n-1,1}&\cdots&A_{n-1,n-1}&0\\ A_{n1}&\cdots&A_{n,n-1}&I_m \end{bmatrix}\in M_{n-1}(R) $$ and $Det(B)$ is not a zero divisor. It follows that $$ \left[ \begin{array}{ccc|c} I_m&&&0\\ &\ddots&&\vdots\\ &&I_m&0\\ \hline (-1)^{n+1}Det(\widehat{A}_{n1})&\cdots&-Det(\widehat{A}_{n,n-1})&Det(\widehat{A}_{nn}) \end{array}\right] \,M =\begin{bmatrix}\widehat{A}_{nn}&\ast\\ 0&L_{\ast n}(M). \end{bmatrix} $$ Take determinants on both sides, we get $$ \det(Det(\widehat{A}_{nn}))\det(M)=\det(\widehat{A}_{nn})\det(L_{\ast n}(M)).\tag{3} $$ However, by Silvester, if all sub-blocks of a block matrix $B$ commute, then $\det(Det(B))=\det(B)$. Therefore $(3)$ reduces to $\det(\widehat{A}_{nn})\det(M)=\det(\widehat{A}_{nn})\det(L_{\ast n}(M))$. Yet, $det(\widehat{A}_{nn})$ is not a zero divisor. Therefore we may cancel it out on both sides and get $\det(M)=\det(L_{\ast n}(M))$.

Answer 2

As an alternative to Laplace Expansion one may proceed in this case by Block Gauss Elimination as follows.

Perform the block row operations:

$$R_4:=R_4+XR_5;\ R_3:=R_3+XR_4;\ R_2:=R_2+XR_3;\ R_1:=R_1+XR_2 $$ which replaces the $X$s on the first four diagonal entries by $O$s and replaces the $V_0$ term by $X^5+V_4 X^4+V_3 X^3 + V_2 X^2 + V_1 X +V_0$, and the other $V_i$ by corresponding terms.

Now perform a block column operation, $$ C_5:=C_5+\left(X^4+V_4 X^3+V_3 X^2 + V_2 X^1 + V_1\right)C_1+\dots+\left(X +V_4 \right)C_4 $$ to replace all the entries in the last column except the first by $O$.

Now block re-order the rows to obtain the determinant as $$ \det \begin{pmatrix} -I &O &O &O &O \\ O &-I &O &O &O \\ O &O &-I& O &O \\ O &O &O &-I&O \\ O &O &O &O &X^5+V_4 X^4+V_3 X^3 + V_2 X^2 + V_1 X +V_0\\ \end{pmatrix} $$ which evaluates to $$\left(\det(-I)\right)^4 \det\left(X^5+V_4 X^4+V_3 X^3 + V_2 X^2 + V_1 X +V_0\right)=\det\left(X^5+V_4 X^4+V_3 X^3 + V_2 X^2 + V_1 X +V_0\right). $$