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# 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

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{2}=0 .$ Find the determinant of $I+A$ where I denotes the identity matrix.

Determinant

Eigen Values

Eigen Vectors

## 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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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

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{2}=0 .$ Find the determinant of $I+A$ where I denotes the identity matrix.

Determinant

Eigen Values

Eigen Vectors

## 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 .

## Subscribe to Cheenta at Youtube

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