Tuesday, December 6, 2011

More Linear Algebra: Eigenvectors

In the previous instalments on linear algebra, we saw how one can obtain the characteristic polynomial from a matrix as well as the eigenvalues. We now want to use past examples to show how the eigenvectors can be obtained. Consider again the matrix given in the first problem:

A =

(1.....i)
(-i.....1)

for which we found the eigenvalues were:

E1 = 0, E2 = 2, viz.

http://brane-space.blogspot.com/2011/12/solutions-to-linear-algebra-problems.html

To get the eigenvectors is just straightforward and merely requires obtaining simultaneous equations in x, y for example - based on using the rows in the matrix and applying each of the eigenvalues to them.

For example, take E1 = 0, then the resulting equations are:

x - iy = 0

ix + y = 0

or x = -iy and y = ix

The eigenvector is easily solved for and is:

v1 =
[-i]
[1]

Next, take the eigenvalue E2 = 2, then the simultaneous equations from the matrix A are:

x + iy = 2x

-ix + y = 2y

which yields: x = iy and y = -ix

Or, an eigenvector of: v2 =

[1]
[-i]

Consider now problem (2) which featured,

A =
(1.. …. .2)
(2.......-2)

And for which we found: E1 = -3, and E2 = 2

To get the eigenvector associated with the eigenvalue E1 = -3, we form the left side of the algebraic equations in x, and y using A such that:

x + 2 y = -3x

2x - 2y = -3y

Simplifying:

4x + 2y = 0

2x + y = 0

or:

4v1 + 2v2 = 0
2v1 + v2 = 0

and solving the simultaneous eqn. yields: v2 = - 2v1, or

v =

[1]
[-2]

For E2 = 2, we may write:

x + 2y = 2x

2x -2y = 2y

Again, the eigenvalue (E2) is always multiplied by x and then y to give the column matrix comprising the right side, with x-value on top, and y-value on the bottom. Simplifying the preceding equations:

-x + 2y = 0

2x - 4y = 0

Or:

-v1 + 2v2 = 0

2v1 - 4v2 = 0

which yields: v1 = 2v2, so the eigenvector in this case is:

v =

[2]
[1]

Problems:

1) Find the eigenvectors associated with the matrix A in Problem (3), e.g. for

A =

(3.. ……2)
(-2...... 3)

with the eigenvalues: E1 = 3 + 2i, and E2 = 3 - 2i

(2) Given the matrix:

A =

(2.......4)
(5.......3)

Find: a) the characteristic polynomial, b) the eigenvalues associated with it, and c) the eigenvectors.