I'm reading the book Polarimetric Radar Imaging: From basics to applications
On page 75, there's this claim:
$$T_3 = \begin{bmatrix}
2A_0 & C-jD & H+jG\\
C+jD & B_0+B & E+jF\\
H-jG & E-jF & B_0-B
\end{bmatrix}$$
As the coherency $T_3$ matrix in such a case is a rank 1 Hermitian matrix, it follows
that its nine principal minors are zero, with:
$$\bbox[yellow]{2A_0(B_0+B)-C^2-D^2=0}\\
\bbox[yellow]{2A_0(B_0-B)-G^2-H^2=0}\\
-2A_0E+CH-DG=0\\
\bbox[yellow]{B_0^2-B^2-E^2-F^2=0}\\
C(B_0-B)-EH-GF=0\\
-D(B_0-B)+FH-GE=0\\
2A_0F-CG-DH=0\\
-G(B_0+B)+FC-ED=0\\
H(B_0+B)-CE-DF=0$$
I am wondering because I know that a $3\times 3$ matrix has nine $2\times 2$ submatrices and so nine minors, but only 3 of them (the highlighted ones are principal minors)?!!
The others are only submatrices (not principals) and so the other six equations are resulted from $\Re{|A_{({1,3},{1,2})}|}=0$,$\Im{|A_{({1,3},{1,2})}|}=0$,$\Re{|A_{({2,3},{1,2})}|}=0$,$\Im{|A_{({2,3},{1,2})}|}=0$,$\Re{|A_{({2,3},{1,3})}|}=0$ , $\Im{|A_{({2,3},{1,3})}|}=0$
Knowing that when a $3\times 3$ matrix is of rank 1, the determinant of all of its $2\times 2$ submatrices ($2\times 2$ minors) should be zero not just the principal ones.
Best Answer
For a general $3\times 3$ matrix, $$A =\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$ there is one third order principal minor namely $|A|$.
There are three second order principal minors:
$\begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix}$ formed by deleting column $3$ and row $3$.
$\begin{vmatrix} a_{11} & a_{13}\\ a_{31} & a_{33} \end{vmatrix}$ formed by deleting column $2$ and row $2$.
$\begin{vmatrix} a_{22} & a_{23}\\ a_{32} & a_{33} \end{vmatrix}$ formed by deleting column $1$ and row $1$.
There are three first order principal minors: $|a_{11}|, |a_{22}|, a_{33}|$ .