Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More

# ISI MStat PSB 2005 Problem 3 | The Orthogonal Matrix

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.

### Prerequisites

Orthogonal Matrix

Eigenvalues

Characteristic Polynomial

## Solution :

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 !!

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.