Tuesday, April 25, 2017

Math Revisited: Linear Algebra (2)

We saw how one can obtain the characteristic polynomial from a matrix as well as the eigenvalues. We now want to use similar methods to show how the eigenvectors can be obtained.   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.

Consider for example:

A =

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

We have P_A(t) =

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

So: P_A(t) = (t - 1) 2  - (-i)(i) = t 2 -2t +1 -1 = 0

Then: P_A(t)   = t 2  - 2t = t(t - 2)

The eigenvalues E1,2 are:

E1 = 0, E2 = 2

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]


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 =


[-i]

Consider now:

A =

(1.. …. .2)
(2.......-2)

Then:  P_A(t) =

(t - 1......2)
2........t +2)

P_A(t) = (t - 1)((t + 2) - 4

P_A(t) = t  2 + t - 2  -4 =  t 2 + t - 6

But: t 2 + t - 6 = (t + 3) (t - 2)

So: 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 =


[-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 =




Working these linear algebra problems is fairly straightforward once one follows the steps such as I've outlined above.