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# 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!

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 !!

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