Difference between revisions of "Linear algebra (eigenvalues and eigenvectors)" - Rhea

Revision as of 06:11, 6 December 2010

1) Definition
• Let
              L:V → V

be the linear transformation of n-dimensional vector space into itself (a linear operator on V)
Then, λ is an eigenvalue of L if there exists a non-zero vector x in V such that
          L(x) = λx

In other words, λ is a scalar associated with vector x to represent a linear transformation.

 • If λ = eigenvalue, then x = eigenvector (an eigenvector is always associated with an eigenvalue)
Eg: If L(x) = 5x , 5 is the eigenvalue and x is the eigenvector.
* λ can be either real or complex, as will be shown later.

• An example about the concept of eigenvalue and eigenvector, based on a linear transformation:

(Read the matrix [a , b] as entry [a] in the first row and entry [b] in the second row)

• Given that L:R² → R² be linear operator defined by
L([a₁ , a₂]) = [-a₂ , a₁]
To find the eigenvalues of L and the associated eigenvectors, we have to find λ such that
                   L([a₁ , a₂]) = λ[a₁, a₂]

Since [-a₂ , a₁] = λ[a₁ , a₂], equate them together, and we can find that 
                   λa₁ = -a₂  (1) and λa₂ = a₁ (2)

By substituting (2) into (1), we obtain
λ(λa₂) = -a₂
λ²a₂ = - a₂
λ² = -1
Since the square root of -1 is equal to the complex numberi, we can conclude that λ = ±i
Because the eigenvalues are not real, we can say that there is no vector [a₁ , a₂] in R² such that L([a₁ , a₂]) is parallel to [a₁ , a₂].

But if L is mapped from C² into C² , then L has eigenvalue i (eigenvector [i , 1]) and eigenvalue -i (eigenvector [-i , 1]

* Zero vector cannot be an eigenvector, but scalar zero can be eigenvalue* - IMPORTANT FACT

• An example about the concept of eigenvalue and eigenvector, based on a linear transformation AND basis:

• Given that L:P₂ → P₂ be linear operator defined by
    L(at² + bt + c) = -bt-2c

Question : Find the corresponding matrix eigen-problem for basis S = [1-t, 1+t, t²].

Solution:

TIME FOR SOME MATRICES !!!

Before even attempting to find eigenvalues and associated eigenvectors for matrices, two things we should know: CHARACTERISTIC POLYNOMIAL OF A MATRIX and CHARACTERISTIC EQUATION OF A MATRIX.

1. CHARACTERISTIC POLYNOMIAL OF A MATRIX

To find the eigenvalues and associated eigenvectors of a matrix, we should construct a *NEW* matrix out of a given matrix. The *NEW* matrix is where we can get the characteristic polynomial of the given matrix from. 

The significance of the characteristic polynomial of a matrix : the roots of the polynomial are the eigenvalues of the given matrix (Theorem 7.1), and by substituting eigenvalues into the characteristic polynomial and solving for the solution, we can obtain the associated eigenvectors for the eigenvalues of the matrix

    2.  CHARACTERISTIC EQUATION OF A MATRIX

In order to find the eigenvalues and associated eigenvectors of a given matrix, we would need to have a particular   equation.  The equation could be obtained  by using the concept of determinants.

.The equation : p(λ) = det(λI_n - A) = 0, where λ is the variable that represents the eigenvalues, I_n = Identity matrix and A = the given matrix

HOW TO CONSTRUCT A CHARACTERISTIC POLYNOMIAL (A SHORTCUT):

(Read the matrix [a b , c d] as entries [a] and [b] in the first row, and entries [c] and [d] in the second row)

Let's say the question requires us to solve for the eigenvalues and associated eigenvectors of a nXn matrix A,

  [1 1 , -2 4]

Step 1 :To come up with the characteristic polynomial of the matrix, change the entries on the main diagonal to from [x] to [λ-x] (or from [-x] to [λ+x]), and change the sign of the other entries of the matrix (if the entries are zero on the diagonal, just change to λ, but if the other entries are zero, just remain as zero)

For the question asked above, the characteristic polynomial would be [λ-1 -1 , 2 λ-4] (Note the entries of the main diagonal have a λ sign, and the other entries have changed signs).

Step 2 : Using the concept of determinants learnt in Chapter 3, compute the determinant of the matrix (Note that we are going to get an algebraic equation, so be careful when doing the computation, a small mistake might result in a wrong answer)

Using the concept of det, we can compute the characteristic polynomial of the matrix. To refresh our minds,

det (A) = (λ-1)(λ-4)-(-2) = 0

λ²-5λ+6 = 0

(λ-2)(λ-3) = 0

Hence the roots of the characteristic polynomial @ the eigenvalues of the matrix are λ=2 and λ=3.

Step 3 : To compute the eigenvectors associated with the eigenvalues, there are two ways: (1) using the equation A(x) = λ(x), where A = the given matrix, x = the solution (eigenvector) and λ = eigenvalue or (2) using the eigenvalue obtained, just substitute it back into the *NEW* matrix and solve for the solution (eigenvector) by doing reducing the matrix to reduced row echelon form (rref).

For this question, let's try to solve for the associated eigenvectors by using the second way, i.e substituting the eigenvalue into the *NEW* matrix and doing rref.

When λ = 2, we have the matrix [1 -1 , 2 -2] [x₁ , x₂]. After reducing the matrix to rref, we have [1 -1 , 0 0], hence indicating that the matrix have a nontrivial solution.

Choosing the second column as our independent variable (because there is no leading one), we have x₂ = r, and x₁ - r = 0, so x₁ = r.

Hence we have [r , r] as our solution, and by factorizing r out of the solution space, we have eigenvector [1 , 1] for λ=2

Repeating the same method to find the eigenvector when λ=3, we have the matrix [2 -1, 2 -1][x₁ , x₂]. After reducing the matrix to rref, we have [1 -1/2 , 0 0], hence indicating that the matrix have a nontrivial solution.
Choosing the second column as our independent variable (because there is no leading one), we have x₂ = r, and x₁ - r/2 = 0, so x₁ = r/2.

Hence we have [r/2 , r] as our solution, and by factorizing r out of the solution space, we have eigenvector [1/2 , 1] for λ=3.

SOLVED!!!