+2354 Can someone help me with this thing, It has been a while since I worked with complex values.
Given the following matrix; (2−510)
I had to find the eigenvalues.
since det(A-uI) = 0 for eigenvalues u; det(2−u−510−u)=0⇒u2−2u+5=0⇒u=2±√4−4∗52
So
u1=1+2i
u2=1−2i
Now find the eigenvectors.
I know I need to solve
\begin{pmatrix} 2 & -5\\ 1 & 0 \end{pmatrix} \begin{pmatrix} v_1_1 \\ v_1_2 \end{pmatrix} = \begin{pmatrix} 1+2i \end{pmatrix} \begin{pmatrix} v_1_1 \\ v_1_2 \end{pmatrix}
and
\begin{pmatrix} 2 & -5\\ 1 & 0 \end{pmatrix} \begin{pmatrix} v_1_1 \\ v_1_2 \end{pmatrix} = \begin{pmatrix} 1-2i \end{pmatrix} \begin{pmatrix} v_1_1 \\ v_1_2 \end{pmatrix}
but I get stuck trying to solve it.
Reinout 
+6252 λ1=1+2ı
[2−λ1−51−λ1]⇒[1−2ı−51−1−2ı]⇒
[1−1−2ı1−2ı−5]⇒[1−1−2ı00]
x−(1+2ı)y=0
v1=[11+2ı]
λ2=1−2ı
[2−λ2−51−λ2]⇒[1+2ı−51−1+2ı]⇒
[1−1+2ı1+2ı−5]⇒[1−1+2ı00]
x−(1−2ı)y=0
v2=[11−2ı]
+130477 Mmmm....let's look at this one again....
We need to solve
I)u=0 
1 -2i -5 x = 0
1 -1 - 2i y = 0
This generates the following system
(1 - 2i) x - (5)y = 0
1x -(1 + 2i)y = 0
Using the second equation, we have
x = (1 + 2i)y
And if we let y = some convenient value, say, 1....then x = (1 + 2i) and we have the following eigenvector associated with lambda = (1 + 2i)
(1 + 2i)
1
BTW....you can check this solution in the 1st equation of the system.....I believe it "works"
To find the eigenvector associated with the second lambda, (1 - 2i), we follow a similar process and we have
1 +2i -5 x = 0
1 -1 + 2i y = 0
This generates the following system
(1 + 2i) x - (5)y = 0
1x -1y + 2iy = 0
And using the second equation, we have
x = (1 - 2i)y
And if we let y = some convenient value, say, 1....then x = (1 - 2i) and we have the following eigenvector associated with lambda = (1 - 2i)
(1 - 2i)
1
As before....you can check this solution in the 1st equation of the system.....
I think this is correct, but my Linear Algebra skills have been sitting in the closet for awhile...
