## Thursday, December 8, 2011

### Solutions to Linear Algebra Problems (Eigenvectors)

We left off with two problems on obtaining eigenvectors from matrices:

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
--

Solution:

We proceed by writing out the algebraic equations with the eigenvalues:

3x + 2y = (3 + 2i) x = 3x + 2i

-2x + 3y = (3 +2i)y = 3y + 2iy

Which, on simplification, reduces to: 2y = 2ix

Or: y = ix

And for the bottom eqn.: -2x = 2iy or -x = iy and x = -iy

Then, v =

[i]
[-i]

For the second eigenvalue (3 – 2i) we can write the equations:

3x + 2y = (3 – 2i)x = 3x – 2ix

-2x + 3y = (3 – 2i)y = 3y – 2iy

Simplifying each line:

2y = - 2ix

-2x = - 2iy

Or: y = -ix and x = iy

So: v =

[-i]
[i]

(2) Given the matrix:A =

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

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

Solution:

a) We write: P_A(t) =

(t – 2………4)
(5………..t – 3)

Then: P_A(t) = (t – 2)(t – 3) – 4(5) = t^2 – 5t + 6 – 20 = 0

Or: t^2 – 5t – 14 = 0

b) Factoring the polynomial P_A(t):

P_A(t) = ( t – 7)( t + 2)

We solve for the eigenvalues E2, E2, where t1 = E1 and t2 = E2.

Then: E1 = 7 and E2 = -2

c) The Eigenvectors:

First, take E1 and write its algebraic equation from the matrix:

2x + 4y = E1(x) = 7x

5x + 3y = E1 (y) = 7y

Simplifying the respective lines:

-5x + 4y = 0

5x – 4y = 0

Or in vector form:

-5v1 + 4v2 = 0

5v1 – 4v2 = 0

Where v1 = 4v2/5 = 0.8 v2

So: v =

[0.8]
[1.0]

The second eigenvalue is E2 = -2 so we have:

2x + 4y = E2(x) = -2x

5x + 3y = E2 (y) = -2y

And after simplifying:

4x + 4y = 0

5x + 5y =0

For which v1 = v2 and

V =
[1]
[1]