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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!