This is a very subtle sample problem from ISI MStat PSB 2005 Problem 3. Given that one knows the property of orthogonal matrices its just a counting problem. Give it a thought!
Problem– ISI MStat PSB 2005 Problem 3
Let \(A\) be a \(n \times n\) orthogonal matrix, where \(n\) is even and suppose \(|A|=-1\), where \(|A|\) denotes the determinant of \(A\). Show that \(|I-A|=0\), where \(I\) denotes the \(n \times n\) identity matrix.
This is a very simple problem, when you are aware of the basic facts.
We, know that, the eigenvalues of a orthogonal matrix is \(-1\) and \(1\) .(\(i\) and \(-i\) if its skew-symmetric). But this given matrix \(A\) is not skew-symmetric.(Why??).So let for the matrix \(A\), the algebraic multiplicity of \(-1\) and \(1\) be \(m\) and \(n\), respectively.
So, since \(|A|=-1\), hence the algebraic multiplicity of \(-1\) is definitely odd, since we know by the property of eigenvalues determinant of a matrix is just the product of its eigenvalues.
Now since, \(n\) is even and the algebraic multiplicity of \(-1\) i.e. \(m\) is odd, hence \(n\) is also odd and \(n \ge 1\).
Hence, the Characteristic Polynomial of \(A\), is \(|I\lambda – A |=0\), where \(\lambda\) is the eigenvalue of \(A\), and in this problem \(\lambda=-1 \) or \( 1\).
Hence, putting \(\lambda=1\), we conclude that, \(|I-A|=0\). Hence we are done !!
Food For Thought
Now, suppose \(M\) is any non-singular matrix, such that \(M^2=-I\). What can you say about the column space of \(M\) ?
Keep thinking !!