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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!

## Problem- ISI MStat PSB 2005 Problem 3

Let be a orthogonal matrix, where is even and suppose , where denotes the determinant of . Show that , where denotes the 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 and .( and if its skew-symmetric). But this given matrix is not skew-symmetric.(Why??).So let for the matrix , the algebraic multiplicity of and be and , respectively.

So, since , hence the algebraic multiplicity of is definitely odd, since we know by the property of eigenvalues determinant of a matrix is just the product of its eigenvalues.

Now since, is even and the algebraic multiplicity of i.e. is odd, hence is also odd and .

Hence, the Characteristic Polynomial of , is , where is the eigenvalue of , and in this problem or .

Hence, putting , we conclude that, . Hence we are done !!

## Food For Thought

Now, suppose is any non-singular matrix, such that . What can you say about the column space of ?

Keep thinking !!

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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 be a orthogonal matrix, where is even and suppose , where denotes the determinant of . Show that , where denotes the 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 and .( and if its skew-symmetric). But this given matrix is not skew-symmetric.(Why??).So let for the matrix , the algebraic multiplicity of and be and , respectively.

So, since , hence the algebraic multiplicity of is definitely odd, since we know by the property of eigenvalues determinant of a matrix is just the product of its eigenvalues.

Now since, is even and the algebraic multiplicity of i.e. is odd, hence is also odd and .

Hence, the Characteristic Polynomial of , is , where is the eigenvalue of , and in this problem or .

Hence, putting , we conclude that, . Hence we are done !!

## Food For Thought

Now, suppose is any non-singular matrix, such that . What can you say about the column space of ?

Keep thinking !!

## Subscribe to Cheenta at Youtube

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