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Eigen Values | ISI MStat 2019 PSB Problem 2

This post based on eigen values of matrices and using very basic inequalities gives a detailed solution to ISI M.Stat 2019 PSB Problem 2.

Problem

Let A and B be 4 \times 4 matrices. Suppose that A has eigenvalues x_{1}, x_{2}, x_{3}, x_{4} and B has eigenvalues \frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}}, \frac{1}{x_{4}} where each x_{i}>1
(a) Prove that A+B has at least one eigenvalue greater than 2.
(b) Prove that A-B has at least one eigenvalue greater than 0.
(c) Give an example of A and B so that 1 is not an eigenvalue of AB.

Prerequisites

  • Trace of a Matrix
  • AM - GM Inequality
  • Pigeon Hole Principle: Let the average of a set of positive numbers be \mu. Use the pigeonhole principle to show that there exists at least one number less than or equal to \mu.
  • Algebra of Diagonal Matrices
  • Eigenvalues of a Diagonal Matrix are the diagonal elements.

Solution

(a)

Consider the trace of A+B.

Mean of the eigen values of A + B = \frac{Tr(A+B)}{4} = \frac{Tr(A)+ Tr(B)}{4} = \frac{x_{1} +  x_{2} + x_{3} + x_{4}+ \frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} + \frac{1}{x_{4}}}{4} = \frac{\sum_{i = 1}^{4}(x_{i} + \frac{1}{x_{i}})}{4} \overset{x_{i} + \frac{1}{x_{i}} \geq 2}{\geq} 2. [ Hint: AM - GM Inequality ].

Now, use Pegion Hole Principle as mentioned in the prerequisites.

(b)

Mean of the eigen values of A - B = \frac{Tr(A-B)}{4} = \frac{Tr(A) - Tr(B)}{4} = \frac{x_{1} + x_{2} + x_{3} + x_{4} - \frac{1}{x_{1}} - \frac{1}{x_{2}} - \frac{1}{x_{3}} - \frac{1}{x_{4}}}{4} = \frac{\sum_{i = 1}^{4}(x_{i} - \frac{1}{x_{i}})}{4} \overset{x_{i} - \frac{1}{x_{i}} > 0}{>} 0. [ Hint: a >  1 \Rightarrow a^2 > 1.]

Now, use Pegion Hole Principle as mentioned in the prerequisites.

(c)

Let's take A = diag( 2, 3, 4, 5) and B  = diag( \frac{1}{2} , \frac{1}{3} , \frac{1}{4} . \frac{1}{5} ). Now observe that A and B satisfy the given conditions.

AB = I. But I has an eigenvalue 1.

So, what to do? We want none of the diagonal values of AB to be not 1.

Take, A = diag( 2, 3, 4, 5) and B = diag( \frac{1}{3} , \frac{1}{2} , \frac{1}{5} , \frac{1}{4} ). ow observe that A and B satisfy the given conditions.

AB = diag( \frac{2}{3} , \frac{3}{2} , \frac{4}{5} . \frac{5}{4} ), which has no eigen value 1.

Stay tuned!

This post based on eigen values of matrices and using very basic inequalities gives a detailed solution to ISI M.Stat 2019 PSB Problem 2.

Problem

Let A and B be 4 \times 4 matrices. Suppose that A has eigenvalues x_{1}, x_{2}, x_{3}, x_{4} and B has eigenvalues \frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}}, \frac{1}{x_{4}} where each x_{i}>1
(a) Prove that A+B has at least one eigenvalue greater than 2.
(b) Prove that A-B has at least one eigenvalue greater than 0.
(c) Give an example of A and B so that 1 is not an eigenvalue of AB.

Prerequisites

  • Trace of a Matrix
  • AM - GM Inequality
  • Pigeon Hole Principle: Let the average of a set of positive numbers be \mu. Use the pigeonhole principle to show that there exists at least one number less than or equal to \mu.
  • Algebra of Diagonal Matrices
  • Eigenvalues of a Diagonal Matrix are the diagonal elements.

Solution

(a)

Consider the trace of A+B.

Mean of the eigen values of A + B = \frac{Tr(A+B)}{4} = \frac{Tr(A)+ Tr(B)}{4} = \frac{x_{1} +  x_{2} + x_{3} + x_{4}+ \frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} + \frac{1}{x_{4}}}{4} = \frac{\sum_{i = 1}^{4}(x_{i} + \frac{1}{x_{i}})}{4} \overset{x_{i} + \frac{1}{x_{i}} \geq 2}{\geq} 2. [ Hint: AM - GM Inequality ].

Now, use Pegion Hole Principle as mentioned in the prerequisites.

(b)

Mean of the eigen values of A - B = \frac{Tr(A-B)}{4} = \frac{Tr(A) - Tr(B)}{4} = \frac{x_{1} + x_{2} + x_{3} + x_{4} - \frac{1}{x_{1}} - \frac{1}{x_{2}} - \frac{1}{x_{3}} - \frac{1}{x_{4}}}{4} = \frac{\sum_{i = 1}^{4}(x_{i} - \frac{1}{x_{i}})}{4} \overset{x_{i} - \frac{1}{x_{i}} > 0}{>} 0. [ Hint: a >  1 \Rightarrow a^2 > 1.]

Now, use Pegion Hole Principle as mentioned in the prerequisites.

(c)

Let's take A = diag( 2, 3, 4, 5) and B  = diag( \frac{1}{2} , \frac{1}{3} , \frac{1}{4} . \frac{1}{5} ). Now observe that A and B satisfy the given conditions.

AB = I. But I has an eigenvalue 1.

So, what to do? We want none of the diagonal values of AB to be not 1.

Take, A = diag( 2, 3, 4, 5) and B = diag( \frac{1}{3} , \frac{1}{2} , \frac{1}{5} , \frac{1}{4} ). ow observe that A and B satisfy the given conditions.

AB = diag( \frac{2}{3} , \frac{3}{2} , \frac{4}{5} . \frac{5}{4} ), which has no eigen value 1.

Stay tuned!

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One comment on “Eigen Values | ISI MStat 2019 PSB Problem 2”

  1. We have first prove the eigenvalues of A+B and A-B are real. Also, there are examples online contradicting the first to statements.

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