Question

Let A be (2n+1)×(2n+1) matrix with integer entries and positive determinant,where n∈N.If AA^T = I = A^TA, then which of the following statements always holds?

• A. det(A)=0
• B. det(A+I)≠0
• C. det(A+I)=0
• D. det(A-I)=0
• E. det(A-I)≠0

Answer 1

The Correct Option is D

Given \( A \) is a \((2n+1) \times (2n+1)\) integer matrix with \( AA^T = I = A^TA \) and \(\det(A) > 0\). 

Since \( AA^T = I \), \( A \) is an orthogonal matrix. For orthogonal matrices: \[ \det(A) = \pm 1 \] Given \(\det(A) > 0\), we have \(\det(A) = 1\). 

For any orthogonal matrix \( A \) of odd dimension, \( 1 \) is always an eigenvalue. This is because: \[ \det(A - I) = \det(A^T - I) = \det(A^{-1} - I) = \det(A)^{-1}\det(I - A) = (-1)^{2n+1}\det(A - I) \] For odd \( 2n+1 \), this implies: \[ \det(A - I) = -\det(A - I) \implies \det(A - I) = 0 \] 

Answer 2

Given: \(A\) is a \((2n+1) \times (2n+1)\) matrix with integer entries, \(\det(A) > 0\), and \(AA^T = I = A^TA\).

This means \(A\) is an orthogonal matrix and \(\det(A) = \pm 1\). Since \(\det(A) > 0\), we have \(\det(A) = 1\).

Consider \(A - I\). We want to determine if \(\det(A - I) = 0\) or \(\neq 0\).

Since \(A\) is a real orthogonal matrix of odd size \((2n+1) \times (2n+1)\), it has at least one real eigenvalue equal to 1. 

This is because the complex eigenvalues of a real matrix come in conjugate pairs, and the product of all eigenvalues must be \(\det(A) = 1\). 

To get a positive determinant from a product containing complex conjugate pairs(each of which multiply to a positive real), there must be at least one additional real eigenvalue, which must be \(1\) since \(A\) is orthogonal.

If \(\lambda = 1\) is an eigenvalue of \(A\), then \(\det(A - I) = 0\).

Therefore, the correct answer is:

(D) \(\det(A - I) = 0\)