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I am trying to find out under what circumstances a matrix of this form would have a zero and/or non-zero determinant:

Lets say I have a matrix of this form; where the sub-matrix from $a_{55}=1; a_{66}=1; a_{77}=1 \dots a_{nn}=1$, and all of the other matrix elements with j and k both greater than 4 have $a_{jk}=0$

\begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} & & & & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & a_{24} & & & & \dots & a_{2n} \\ a_{31} & a_{32} & a_{33} & a_{34} & & & & \dots & a_{3n} \\ a_{41} & a_{42} & a_{43} & a_{44} & & & & \dots & a_{4n} \\ a_{51} & a_{52} & a_{53} & a_{54} & 1 & 0 & 0 &\dots & 0 \\ a_{61} & a_{62} & a_{63} & a_{64} & 0 & 1 & 0 &\dots & 0 \\ a_{71} & a_{72} & a_{73} & a_{74} & 0 & 0 & 1 &\dots & 0 \\ \vdots & & \vdots & & & \vdots & & & \vdots \\ a_{n1} & a_{n2} & a_{n3} & a_{n4} & 0 & 0 & 0 &\dots & 1 \\ \end{vmatrix}

Can I write an equation for the determinant of the larger matrix in terms of the determinant of the sub-matrices?

\begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{vmatrix}

The other square sub-matrix from $a_{55} \dots a_{nn}$ trivially has a determinant of 1. Lets further saw I know that these coefficients are small \begin{vmatrix} a_{15} & a_{16} & \dots & a_{1n} \\ a_{25} & a_{26} & \dots & a_{2n} \\ a_{35} & a_{36} & \dots & a_{3n} \\ a_{45} & a_{46} & \dots & a_{4n} \\ \end{vmatrix}

As well, most of these coefficients are also small. \begin{vmatrix} a_{51} & a_{52} & a_{53} & a_{54} \\ a_{61} & a_{62} & a_{63} & a_{64} \\ a_{71} & a_{72} & a_{73} & a_{74} \\ \vdots & & \vdots \\ a_{n1} & a_{n2} & a_{n3} & a_{n4} \\ \end{vmatrix}

I want to prove the full nxn matrix has a non-zero determinant, so I need an equation for the determinant of the nxn matrix in terms of the sub matrices.

Sheldon L
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1 Answers1

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We can compute the determinant of block matrices as follows. $$ \begin{align*} \det \begin{pmatrix} A & B \\ C & D \end{pmatrix} &= \det \begin{pmatrix} A & B \\ 0 & D - CA^{-1}B \end{pmatrix} \\ \\ &= \underbrace{\det(A)\cdot \det\left(D-CA^{-1}B\right)}_\text{if $A^{-1}$ exists} \\ \\ &= \underbrace{\det\left(AD-CB\right)}_\text{if $AC=CA$} & \end{align*} $$ In your case, we have $A=I_n$ (and interchanging the diagonal blocks, for an analogous formula with $D=I_n$).

Reference: Proofs of Determinants of Block matrices.

Dietrich Burde
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