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 !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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 !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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