ISI MStat PSB 2007 Problem 1 | Determinant and Eigenvalues of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 1 based on Determinant and Eigen values and Eigen vectors . Let's give it a try !!

Problem- ISI MStat PSB 2007 Problem 1

Prerequisites

Solution :

Let \( {\lambda}_{1} , {\lambda}_{2} \) be two eigen values of A then , \( {{\lambda}_{1}}^2 , {{\lambda}_{2}}^2 \) .

Now it's given that \( A^2=0 \) , so we have \( {{\lambda}_{1}}^2=0 , {{\lambda}_{2}}^2 =0 \) . You may verify it ! (Hint : use the theorem that \( \lambda \) is a eigen value of matrix B and \( \vec{x}\) is it's corresponding eigen value then we can write \(Bx=\lambda \vec{x} \) or , use \(det(B- \lambda I )=0 \) ).

Hence we have \( {\lambda}_{1} =0 , {\lambda}_{2}=0 \) .

Now , eigen values of Identity matrix I are 1 . So, we can write for eigen value \( \vec{x}\) of (A+I) , \( (A+I) \vec{x}= Ax+I\vec{x}=0+\vec{x}=\vec{x} \).

Thus we get that both the eigen values of (A+1) are 1 . Again we know that determinant of a matrix is product of it's eigen values .

So, we have \(|A+I|=1\).

Do you think this solution is correct ?

If yes , then you are absolutely wrong . The mistake is in assuming A and I has same eigen vectors \( Ax+I\vec{x} \ne \vec{x} \)

Correct Solution

We have shown in first part of wrong solution that A has eigen values 0 . Hence the characteristic polynomial of A can be written as , \( |A- \lambda I|= {\lambda}^2 \) .

Now taking \( \lambda =-1 \) we get \( |A+ I|={(-1)}^2 \implies |A+I|= 1 \) .

Food For Thought

If we are given that \( A^{n} = 0 \) for positive integer n , instead of \( A^2=0 \) then find the same .

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