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 .