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
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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]
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