# Determinants of minors occurring 'within' determinant of full matrix

$$A= (a_{ij})$$ is an $$n\times n$$ symmetric positive matrix. It induces a quadratic form $$f(x):= x^tAx$$ on $$\mathbb{R}^n$$. $$D_m$$ denotes the determinant of the top left $$m\times m$$ submatrix of $$A$$ (or rather of $$f$$). What does the following highlighted sentence mean?

This is from page 781 of Hancock's book on 'Minkowski's geometry of numbers'. It was written in the 1930's and I've been having a hard time with the language.

E.g. When $$A$$ is $$3\times 3$$ and $$m=2$$, we have $$$$D_3 = \det\left[ {\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{array} } \right],\ D_m = D_2 = \det\left[ {\begin{array}{cc} a_{11} & a_{12} \\ a_{12} & a_{22} \end{array} } \right],\ D_{n-m}=D_1 = a_{11}.$$$$

## 2 Answers

If the $$n\times n$$ matrix $$M$$ is decomposed into submatrices, $$M=\begin{pmatrix}A&B\\ C&D\end{pmatrix},$$ where $$A$$ has dimension $$m\times m$$, then the determinant of $$M$$ can be decomposed as $$\det M=\det A\det D+X.$$ The multinomial $$X$$ in the matrix elements of $$M$$ contains $$n!-m!(n-m)!$$ terms, for a general matrix $$M$$. If the matrix is symmetric, the number of distinct terms is less.

In the $$n=3$$, $$m=2$$ example given in the OP, this gives for $$X$$ the four terms $$X=a_{13} a_{22} a_{31} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{11} a_{23} a_{32}.$$ Notice that the indices of $$X$$ follow Hancock's description.

So I would paraphrase the sentence highlighted in yellow as "Write down the determinant $$D_n$$ of $$f$$ and within that expression single out the product of the principal minors $$D_m$$ and $$D_{n-m}$$."

• Thank you! So in the $3\times 3$ example I should be reading $D_{n-m}=D_1$ as $a_{33}$ rather than $a_{11}$? Sorry for confusion. Commented Jan 5, 2022 at 18:14
• Maybe I should read ahead and see if $\overline{D}_{n-m}$ makes more sense than $D_{n-m}$. Commented Jan 5, 2022 at 18:19
• certainly, that is expressed in the quote by "diagonally opposite"; obviously the determinant cannot contain a term $a_{11}(a_{11}a_{22}-a_{12}a_{21})$. Commented Jan 5, 2022 at 19:32
• Thanks, you're right. It is quite clear. Commented Jan 6, 2022 at 2:46
• Sir Beenakker. In the picture above, do you see how the inequality (13) - $2|a_{kh}| \leq a_{kk}$ for $k< h$ implies that 'each of the terms' is less than $\frac{1}{4}a_{11}\dots \widehat{a_{m+1 m+1}}\dots a_{nn}$? Commented Jan 11, 2022 at 10:53

Hancock seems to do not such a good job with the translation of Minkowski's work. I've posted the original here. See for example the double occurrence of $$a_{mm}$$ in the last inequality above versus what's written in the English version.