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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.

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 \(AB\).

- 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.

**(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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