+2353 Can someone help me with this thing, It has been a while since I worked with complex values.
Given the following matrix; $$\begin{pmatrix}
2&-5 \\
1&0
\end{pmatrix}$$
I had to find the eigenvalues.
since det(A-uI) = 0 for eigenvalues u; $$det\begin{pmatrix}
2-u& -5\\
1& 0-u
\end{pmatrix} = 0
\Rightarrow u^2-2u+5 = 0 \Rightarrow u = \frac{2 \pm \sqrt{4-4*5}}{2}$$
So
$$u_1 = 1+2i$$
$$u_2 = 1-2i$$
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 
+6251 $$\lambda_1=1+2\imath$$
$$\begin{bmatrix}2-\lambda_1 &-5 \\ 1 &-\lambda_1 \end{bmatrix} \Rightarrow \begin{bmatrix}1-2\imath &-5 \\ 1 &-1-2\imath\end{bmatrix}\Rightarrow$$
$$\begin{bmatrix}1 &-1-2\imath\\ 1-2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1-2\imath \\0 &0\end{bmatrix}$$
$$x-(1+2\imath)y=0$$
$$v_1=\begin{bmatrix}1 \\1+2\imath\end{bmatrix}$$
$$\lambda_2=1-2\imath$$
$$\begin{bmatrix}2-\lambda_2 &-5 \\ 1 &-\lambda_2 \end{bmatrix} \Rightarrow \begin{bmatrix}1+2\imath &-5 \\ 1 &-1+2\imath\end{bmatrix}\Rightarrow$$
$$\begin{bmatrix}1 &-1+2\imath\\ 1+2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1+2\imath \\0 &0\end{bmatrix}$$
$$x-(1-2\imath)y=0$$
$$v_2=\begin{bmatrix}1 \\1-2\imath\end{bmatrix}$$
.+6251 $$\lambda_1=1+2\imath$$
$$\begin{bmatrix}2-\lambda_1 &-5 \\ 1 &-\lambda_1 \end{bmatrix} \Rightarrow \begin{bmatrix}1-2\imath &-5 \\ 1 &-1-2\imath\end{bmatrix}\Rightarrow$$
$$\begin{bmatrix}1 &-1-2\imath\\ 1-2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1-2\imath \\0 &0\end{bmatrix}$$
$$x-(1+2\imath)y=0$$
$$v_1=\begin{bmatrix}1 \\1+2\imath\end{bmatrix}$$
$$\lambda_2=1-2\imath$$
$$\begin{bmatrix}2-\lambda_2 &-5 \\ 1 &-\lambda_2 \end{bmatrix} \Rightarrow \begin{bmatrix}1+2\imath &-5 \\ 1 &-1+2\imath\end{bmatrix}\Rightarrow$$
$$\begin{bmatrix}1 &-1+2\imath\\ 1+2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1+2\imath \\0 &0\end{bmatrix}$$
$$x-(1-2\imath)y=0$$
$$v_2=\begin{bmatrix}1 \\1-2\imath\end{bmatrix}$$
+129849 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...
