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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 !!

Let be a matrix with real entries such that Find the determinant of where I denotes the identity matrix.

Determinant

Eigen Values

Eigen Vectors

Let be two eigen values of A then , .

Now it's given that , so we have . You may verify it ! (Hint : use the theorem that is a eigen value of matrix B and is it's corresponding eigen value then we can write or , use ).

Hence we have .

Now , eigen values of Identity matrix I are 1 . So, we can write for eigen value of (A+I) , .

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 .

**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

**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 , .

Now taking we get .

If we are given that for positive integer n , instead of then find the same .

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 !!

Let be a matrix with real entries such that Find the determinant of where I denotes the identity matrix.

Determinant

Eigen Values

Eigen Vectors

Let be two eigen values of A then , .

Now it's given that , so we have . You may verify it ! (Hint : use the theorem that is a eigen value of matrix B and is it's corresponding eigen value then we can write or , use ).

Hence we have .

Now , eigen values of Identity matrix I are 1 . So, we can write for eigen value of (A+I) , .

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 .

**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

**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 , .

Now taking we get .

If we are given that for positive integer n , instead of then find the same .

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