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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.

Prerequisites


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 .


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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.

Prerequisites


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 .


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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