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 .

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Eigen Value of a matrix | IIT JAM 2017 | Problem 58

Try this problem from IIT JAM 2017 exam (Problem 58) and know how to evaluate Eigen value of a Matrix.

Eigen value of a Matrix | IIT JAM 2017 | Problem 58

Let $\alpha, \beta, \gamma, \delta$ be the eigenvalues of the matrix
$$
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
Then find the value of $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$

Key Concepts

Linear Algebra

Matrix

Eigen Values

Check the Answer

Answer: $6$

IIT JAM 2017 , Problem 58

Try with Hints

Characteristic Equation of a matrix $A$ of order $n$ is defined by $|A-xI|=0$, where $x$ is scalar and $I$ is the identity matrix of order $n$.

The roots of the characteristic equation are called the Eigen Values of that matrix.

Now it is easy. Give it a try.

Let

$$
A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$

Then its characteristic equation is $|A-xI|=0$ for some scalar $x$ and $I$ the identity matrix of order $4$

i.e., $
\begin{vmatrix}
-x & 0 & 0 & 0 \\
1 & -x & 0 & -2 \\
0 & 1 & -x & 1 \\
0 & 0 & 1 & 2-x
\end{vmatrix}
=0$

$
\Rightarrow -x\begin{vmatrix}
-x & 0 & -2 \\
1 & -x & 1 \\
0 & 1 & 2-x
\end{vmatrix}
=0 $

$\Rightarrow -x[-x\{(-x)(2-x)\}-1]-2(1-0)]=0$

$\Rightarrow x^{4}-2 x^{3}-x^{2}+2 x=0$

$\Rightarrow x(x^{3}-2 x^{2}-x+2)=0$

$\Rightarrow x(x-1)(x-2)(x+1)=0$

Then the eigen values are : $0,1,-1,2$

Then, $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2} =0^2+1^2+(-1)^2+2^2=6$ [ANS]

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