Eigenvalues and Eigenvectors - Mathematical Methods of Physics | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Eigenvalues and eigenvectors

Let A = [ a_ij ] be a given n n  square matrix and consider the equation 

                                                                                  AX = λX                                                                                         ….(1) 

Here X is an unknown vector and  λ an unknown scalar and we want to determine both. A value of  λ for which (1) has a solution X ≠ 0 is called eigenvalue of the matrix A . The corresponding solutions  X ≠ 0 of (1) are called eigenvevtors of A corresponding to that eigenvalue  λ .

In matrix notation,                                    (A - λI)X = 0                                                                                   ….(2)

This homogeneous linear system of equations has a nontrivial solution if and only if the corresponding determinant of the coefficients is zero 

D(λ) is called the characteristic determinant. The equation is called the characteristic equation of the matrix A . By developing D(λ), we obtain a polynomial of n^th degree in λ. This is called the characteristic polynomial of A . 

Note: 

• The eigenvalues of a square matrix A are the roots of the characteristic equation (3) of A . Hence an n x n matrix has at least one eigenvalue and at most n numerically different eigenvalues. 
• Once the eigenvalues are known, corresponding eigenvectors are obtained. 
• Repeated eigenvalues are said to be degenerate eigenvalues. For degenerate eigenvalues there are different eigenvectors for same eigenvalues. 
• Non repeated eigenvalues are non-degenerate eigenvalues. For non-degenerate eigenvalues there are different eigenvectors for different eigenvalues.  
• Sum of eigenvalues are equal to trace of matrix  Trace of matrix is sum of diagonal element. 
• Product of eigenvalues are equal to determinant of matrix
• Eigenvectors correspond to different eigenvalues are always independent.  
• Eigenvectors corresponds to same eigenvalue may or may not be independent.

 Example  5.1:  

Find the eigenvalues and eigenvectors of A= 1 2 2 1 .

Solution:  

det ( A − λ I ) = 1 − λ 2 2 1 − λ = ( 1 − λ ) 2 − 4 = 0

Solutions are λ= 1± 2=− 1, 3.

Now determine the eigenvalues by solving ( A−λ I) e= 0. For λ 1=− 1 we find

2 2 2 2 e 1 e 2 = 0

and thus e 1=− e 2, and e( 1)= c 1 1 . The arbitrary constant can be chosen at will. Some standard choices are 1(simple), 1∕ 2(length 1), etc.

The same algebra for the other eigenvalue leads to e( 2)= d 1− 1 .

If we rewrite

x y = ( x + y ) ∕ 2 1 1 + ( x − y ) ∕ 2 1 − 1 ,

we can easily understand the importantce of the eigenvectors:

A x y = 3 ( x + y ) ∕ 2 1 1 − ( x − y ) ∕ 2 1 − 1 .

Thus the component parallel to ( 1, 1) is stretches by a factor of 3, and the component parallel to (− 1, 1) is inverted (multiplied by − 1)

 A physics example

The most important physical example of the role of the eigenvalue problem can be found in the case of coupled oscillators.

Figure 1 Two coupled oscillators.

Consider Fig.  1 . There we show the case of two masses, coupled by three springs. We assume that x 1 and x 2 are the distances of the masses from the equilibrium position. At that point we assume the strings are untentioned (neither stretched nor compressed).

The equations of motion take a simple form

m 1 ẍ 1 = − k 1 x 1 + k 2 ( x 2 − x 1 ) = − ( k 1 + k 2 ) x 1 + k 2 x 2 m 2 ẍ 2 = − k 3 x 2 − k 2 ( x 2 − x 1 ) = − ( k 3 + k 2 ) x 2 + k 2 x 1

We now take the masses equal ( m 1= m 2= m), and all the spring constants equal as well ( k 1= k 2= k 3= mω 2. We then find that

x ̈ = ω 2 − 2 1 1 − 2 x

This equation can now be solved by writing the standard exponential form, x= e e z t. We then get

z 2 e = ω 2 − 2 1 1 − 2 e

Which is an eigenvalue problem. Write z 2=ω 2λ, and we find that λ=− 1,− 3. Thus z=± iω,± i 3ω. The eigenvectors for these two eigenvalues are ( 1, 1)and ( 1,− 1), respectively.

Thus, in all its generality, we find using superposition that

x = 1 1 ( A cos ( ω t ) + B sin ( ω t ) ) + 1 − 1 ( C cos ( 3 ω t ) + D sin ( 3 ω t ) )                .(4)

This general motion thus consists of the superposition of motion of the two masses in phase ( x 1= x 2, with frequency ω) and one maximally out of phase ( x 1=− x 1, with frequency 3ω).