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Chapter 8: Problem 73

Let \(A\) be any matrix. Show that the signs of a minor \(A[I, J]\) and its complementary minor \(A\left[I^{\prime}, J^{\prime}\right]\) are equal.

Short Answer

Expert verified
In conclusion, the signs of a matrix minor \(A[I, J]\) and its complementary minor \(A[I', J']\) are equal if the size of the original matrix is even. If it's odd, the signs of the minor and its complement will be negatives of each other. This relationship can be expressed through the formula \(\text{sgn}\left(A[I', J']\right) = (-1)^{\left|I\cup J\right|+n}\), and it is proven by considering the number of interchanges of rows and columns between the minor and its complementary minor. This result is important in matrix algebra, particularly in the calculation of inverse matrices.

Step by step solution

01

Understanding Minors and Complementary Minors

First, defining our terms. The minor \( A[I, J] \) is the determinant of a certain square submatrix that can be obtained from the original matrix \( A \) by taking some of its rows (listed in the set \( I \)) and columns (listed in the set \( J \)). Similarly, the complementary minor \( A[I', J'] \) is determined by the rows and columns that are not in the original minor. Thus, \( I' \) and \( J' \) are the complements of \( I \) and \( J \) in the set of all row (or column) indices.
02

Recognizing the relation between a Minor and its Complement

The relationship can be express through formula, given as; \(\text{sgn}\left(A[I', J']\right) = (-1)^{\left|I\cup J\right|+n}\). This formula shows that the sign of the complementary minor of \(A[I, J]\) is equal to that of the original minor times \((-1)^n\), where \(n\) is the size of the matrix.
03

Proving the relation between their signs

This property can be proven by considering permutations of the rows and columns of the matrix. Each interchange of two rows or two columns changes the sign of the determinant. The number of such interchanges needed to go from \(A[I, J]\) to \(A[I', J']\) is the number of elements in the union of \(I\) and \(J\) plus the size of the matrix, \(n\), which is the exponent in the formula. Since every two interchanges bring the determinant to its original sign, the sign of \(A[I', J']\) is equal to that of \(A[I, J]\) times \((-1)^n\).
04

Conclusion

By the count of interchanges (with the help of properties of determinants), it is concluded that the signs of a matrix minor and its complementary minor are indeed identical, assuming the size of the original matrix is even. If it's odd, the signs of the minor and its complement will be negatives of each other. This result is relevant in the calculation of inverse matrices, where the cofactor matrix (the matrix of minors, each multiplied by \((-1)^{i+j}\), where \(i\) and \(j\) are the row and column indices) plays a crucial role.

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Most popular questions from this chapter

Let \(\sigma, \tau \in S_{n} .\) Show that \(\operatorname{sgn}(\tau \circ \sigma)=(\operatorname{sgn} \tau)(\operatorname{sgn} \sigma) .\) Thus, the product of two even or two odd permutations is even, and the product of an odd and an even permutation is odd.

Let \(A\) be an \(n\) -square matrix. Prove \(|k A|=k^{n}|A|\)

Prove Theorem 8.13: Let \(F\) and \(G\) be linear operators on a vector space \(V\). Then (i) \(\operatorname{det}(F \circ G)=\operatorname{det}(F) \operatorname{det}(G)\) (ii) \(F\) is invertible if and only if \(\operatorname{det}(F) \neq 0\)

Prove Theorem \(8.1:\left|A^{T}\right|=|A|\)

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