Eigen Values of a Matrix : IIT JAM 2016 Problem Number 13

What are Eigen Values of a Matrix?

The Eigen values of a matrix are the roots of its characteristic equation.

Try the problem from IIT JAM 2016

The largest eigen value of the matrix

$A = \begin{bmatrix}
1 & 4 & 16 \\[0.3em]
4 & 16 & 1 \\[0.3em]
16 & 1 & 4 \\
\end{bmatrix}$ is

$\textbf{(A)}\qquad 16\qquad \textbf{(B)}\qquad 21\qquad \textbf{(C)}\qquad 48\qquad \textbf{(D)}\qquad 64\qquad $

IIT JAM 2016 Problem Number 13

Finding the largest eigen value of a matrix

6 out of 10

Higher Algebra : S K Mapa

Knowledge Graph- Eigen values of a matrix

Eigen value of matrix knowledge graph

Use some hints

In order to find the largest eigen value of the given matrix we have to find all the eigen values of the given matrix, then we can find the largest among them. Now it is very easy to find the eigen values. Give it a try!!!

Now the process of rank starts with finding the characteristics polynomial of the given matrix, i.e., $\textbf{ det }(A-\lambda I)=0$. We have to find the value of this $\lambda $ which is the eigen value. Now it is very easy to find the determinant of $(A-\lambda I)$. Try to cook this up.

So, now let's do some calculation.

$ \textbf{det}(A-\lambda I) = \left| \begin{matrix}
1-\lambda & 4 & 16 \\[0.3em]
4 & 16-\lambda & 1 \\[0.3em]
16 & 1 & 4 -\lambda \\
\end{matrix} \right| =0 $

Now we are half way done just our calculation part remains.

$|A-\lambda I|=(1-\lambda)[64-20 \lambda + \lambda ^2-1]-4[16-4 \lambda -16]+16[4-256+16 \lambda]=0$

$\Rightarrow \quad (1- \lambda )( \lambda ^2-20 \lambda +63)+16 \lambda +256 \lambda -4032=0$

$\Rightarrow \quad \lambda ^2-20 \lambda +63 - \lambda ^3+20 \lambda ^2-63 \lambda +272 \lambda -4032=0$

$\Rightarrow \quad - \lambda ^3+21 \lambda ^2+189 \lambda -3969=0$

$ \Rightarrow \quad \lambda ^3-21 \lambda ^2-189 \lambda +3969=0 $

$\Rightarrow \quad(\lambda-21)(\lambda^2-189)=0$

Therefore $\lambda=21, \quad -13.747727, \quad 13.747727$

So from here we can clearly see

$\lambda=21$ is the largest among all the eigen values.

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TIFR 2014 Problem 9 Solution - Eigenvalues of Rotation


TIFR 2014 Problem 9 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

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Problem:

Let (A(\theta)=\begin{pmatrix} cos(\theta) &  sin(\theta) \ -sin(\theta) & cos(\theta) \end{pmatrix} )

where (\theta \in (0,2\pi) ). Mark the correct statement below:

A. (A(\theta)) has eigenvectors in (\mathbb{R}^2) for all (\theta \in (0,2\pi))

B.  (A(\theta)) does not have eigenvectors in (\mathbb{R}^2) for any (\theta \in (0,2\pi))

C.  (A(\theta)) has eigenvectors in (\mathbb{R}^2) for exactly one values of  (\theta \in (0,2\pi))

D. (A(\theta)) has eigenvectors in (\mathbb{R}^2) for exactly 2 values of (\theta \in (0,2\pi))

Discussion:

If we try to write the linear operator on (\mathbb{R}^2) which is the clockwise rotation by angle (\theta) then we see that with respect to standard basis, the matrix is (A(\theta)). This is quite easy to see, for ((1,0) \to (cos(\theta),-sin(\theta)) ), and (...) (draw the picture if you are not convinced about this).

What does having an eigenvector mean in this context? Well, as always, it means that (Av=\lambda v) for some (v \neq 0) in (\mathbb{R}^2). But geometrically, this means that upon applying the transformation (in this case rotation) we get the resulting vector in the spanning space of (v), that is the resulting vector must be in the line passing through (v) and the origin.

Geometrically, after rotation, the vector is in the same line is possible only when either the angle of rotation is (0) or the angle of rotation is ( \pi). That is either we did not move at all or we basically reflected about origin.

(\theta \neq 0), so we are left with (\theta=\pi). Therefore, we know option C is true.

If you do not trust your geometric sense at all, then you would want to look at the characteristic polynomial.

We have (det(A(\theta))=cos^2(\theta)+sin^2(\theta)=1) and (trace(A(\theta))=2cos(\theta)).

Therefore, the characteristic polynomial is (x^2-2cos(\theta)x+1). This has a real root if the discriminant is positive. That is if (4cos^2(\theta)-4 \ge 0) i.e, (cos^2(\theta) \ge 1). We know that the maximum value of cos is 1 so (cos(\theta)=1) or (cos(\theta)=-1). Since (\theta \ne 0) we are forced to conclude (\theta = \pi) and we have the answer once more.

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