This is a problem from Schaum's Outlines Linear Algebra.
We are given a n-square matrix $A$ and any $r$ rows-indices $I=\{i_1,i_2,..,i_r\}$ & columns-indices $J=\{j_1,j_2,...,j_r\}$ of $A$ are chosen to define an $r×r$ sub-matrix: $$A(I,J)= [a_{st}: s\in I, t\in J]$$ The determinant $\det(A(I,J))$ is called a minor of $A$ of order $r$ and
$$(-1)^{i_1+i_2+..+i_r+j_1+j_2+..+j_r}\det(A(I,j))$$ is the corresponding signed minor. The complementary minor is $\det(A(I',J'))$ where $I'$ and $J'$ denote the remaining row & column index sets respectively. The theorem says:
The sign of the complementary minor is same as the minor.
Previously this question asked the same but the commenters said it was a false claim. I am a bit confused. A proof would be helpful. Thanks in advance.
It is a false claim. Let $I=J=\{1\}$ and $A=\pmatrix{a&b\\ c&d}$. Then the signed minor for $A(I,J)$ is $(-1)^{1+1}a=a$ and its complementary minor is $(-1)^{2+2}d=d$. Clearly $a$ and $d$ can have different signs.
It is true, however, that the signs multiplied to the minors $\det A(I,J)$ and $\det A(I',J')$ are the same (e.g. in the previous example, we have $(-1)^{1+1}=(-1)^{2+2}$). That is, $(-1)^{\sum_{i\in I}i+\sum_{j\in J}j}=(-1)^{\sum_{i\in I'}i+\sum_{j\in J'}j}$. This is because \begin{aligned} &\left(\sum_{i\in I}i+\sum_{j\in J}j\right)-\left(\sum_{i\in I'}i+\sum_{j\in J'}j\right)\\ &=\left(\sum_{i\in I}i+\sum_{j\in J}j\right)+\left(\sum_{i\in I'}i+\sum_{j\in J'}j\right)-2\left(\sum_{i\in I'}i+\sum_{j\in J'}j\right)\\ &=\left(\sum_{i=1}^ni+\sum_{i=1}^nj\right)-2\left(\sum_{i\in I'}i+\sum_{j\in J'}j\right)\\ &=2\left[\sum_{i=1}^ni-\left(\sum_{i\in I'}i+\sum_{j\in J'}j\right)\right] \end{aligned} is an even number.