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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!