+0  
 
0
764
5
avatar+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 

 Apr 17, 2014

Best Answer 

 #2
avatar+6236 
+10

$$\lambda_1=1+2\imath$$

$$\begin{bmatrix}2-\lambda_1 &-5 \\ 1 &-\lambda_1 \end{bmatrix} \Rightarrow \begin{bmatrix}1-2\imath &-5 \\ 1 &-1-2\imath\end{bmatrix}\Rightarrow$$

$$\begin{bmatrix}1 &-1-2\imath\\ 1-2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1-2\imath \\0 &0\end{bmatrix}$$

$$x-(1+2\imath)y=0$$

$$v_1=\begin{bmatrix}1 \\1+2\imath\end{bmatrix}$$

 

$$\lambda_2=1-2\imath$$

$$\begin{bmatrix}2-\lambda_2 &-5 \\ 1 &-\lambda_2 \end{bmatrix} \Rightarrow \begin{bmatrix}1+2\imath &-5 \\ 1 &-1+2\imath\end{bmatrix}\Rightarrow$$

$$\begin{bmatrix}1 &-1+2\imath\\ 1+2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1+2\imath \\0 &0\end{bmatrix}$$

$$x-(1-2\imath)y=0$$

$$v_2=\begin{bmatrix}1 \\1-2\imath\end{bmatrix}$$

.
 Apr 17, 2014
 #1
avatar+29 
0

the answer is 5

 Apr 17, 2014
 #2
avatar+6236 
+10
Best Answer

$$\lambda_1=1+2\imath$$

$$\begin{bmatrix}2-\lambda_1 &-5 \\ 1 &-\lambda_1 \end{bmatrix} \Rightarrow \begin{bmatrix}1-2\imath &-5 \\ 1 &-1-2\imath\end{bmatrix}\Rightarrow$$

$$\begin{bmatrix}1 &-1-2\imath\\ 1-2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1-2\imath \\0 &0\end{bmatrix}$$

$$x-(1+2\imath)y=0$$

$$v_1=\begin{bmatrix}1 \\1+2\imath\end{bmatrix}$$

 

$$\lambda_2=1-2\imath$$

$$\begin{bmatrix}2-\lambda_2 &-5 \\ 1 &-\lambda_2 \end{bmatrix} \Rightarrow \begin{bmatrix}1+2\imath &-5 \\ 1 &-1+2\imath\end{bmatrix}\Rightarrow$$

$$\begin{bmatrix}1 &-1+2\imath\\ 1+2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1+2\imath \\0 &0\end{bmatrix}$$

$$x-(1-2\imath)y=0$$

$$v_2=\begin{bmatrix}1 \\1-2\imath\end{bmatrix}$$

Rom Apr 17, 2014
 #3
avatar+124595 
+5

Mmmm....let's look at this one again....

We need to solve 

(A− I)u=0
 
I'm going to let x = u 1   and y = u 2.......Also, I don't know how to represent vectors in this thing, but I think you can follow what  I'm doing
 
Going from this point forward, we get

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...

 Apr 17, 2014
 #4
avatar+33151 
+5

Often eigenvectors are normalised to unity.  If you do that here remember, when calculating the normalisation constant, to multiply the complex number component by its complex conjugate; don't just square it!

 Apr 18, 2014
 #5
avatar+2353 
0

Thank you guys!

I highly appreciate your efforts. 

As always, the answer is as clear as daylight once someone gives it to you.

 

 Apr 20, 2014

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