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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 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

Subscribe to Cheenta at Youtube


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

Subscribe to Cheenta at Youtube


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