# Calculating eigenvectors of complex numbers

1. Sep 20, 2009

### boneill3

1. The problem statement, all variables and given/known data

Find the eigenvectors of the matrix A

2. Relevant equations

3. The attempt at the solution

$$\left( \begin{array}{cc}4 & -5 \\1 & 0 \end{array} \right)$$

First I find the characteristic equation $A - \lambda I$

$$\left( \begin{array}{cc}4-\lambda & -5 \\1 & 0-\lambda \end{array} \right)$$

and solve

$\lambda^2-4\lambda+5=0$

which gives the complex eigenvalues $2-i,2+i$

we need to solve

$$\left( \begin{array}{cc}4-(2-i) & -5 \\1 & 0-(2-i) \end{array} \right)$ X = 0$

and

$$\left( \begin{array}{cc}4-(2+i) & -5 \\1 & 0-(2+i) \end{array} \right)$ X = 0$

for the first one I get to

$(2-i)x - 5y = 0$
$x - 2-iy = 0$

I start to get lost after here any help grately appreciated
regards

2. Sep 20, 2009

### CompuChip

So far it looks fine.
Now you have two equations in two unknowns, so you can try to solve them. You will find that they are linearly dependent, so you will only get an expression of the form x = a y.
An eigenvector is then (1, a); the eigenspace is the span of this vector, namely { (x, ax) | x any real number }.

3. Sep 20, 2009

### boneill3

So from the 2 equations I get.

$(2-i)x = 5y$
and
$x = (2-i)y$

so is the first eigen vector

$$\left( \begin{array}{c}1 \\ 2-i \end{array} \right)$$ ?

for the second
$(2+i)x - 5y = 0$
$x - (2+)iy = 0$

I get

$(2+i)x = 5y$
and
$x = (2+i)y$

so is the second eigen vector

$$\left( \begin{array}{c}1 \\ 2+i \end{array} \right)$$ ?

regards
Brendan

4. Sep 20, 2009

### CompuChip

There's an easy way to check your answer... multiply your supposed eigenvectors with the matrix, and check that you get the eigenvector times the corresponding eigenvalue back :)

(Didn't check your calculation, but the result looks good, though)

5. Sep 21, 2009

### boneill3

I multiplied the matrix and the first vector
$$\left( \begin{array}{cc}4 & -5 \\1 & 0 \end{array} \right)$$$$\left( \begin{array}{c}1 \\2-i \end{array} \right)$$

and got

$$\left( \begin{array}{c}-6+5i \\1 \end{array} \right)$$

than Imultiplied the eigenvalue by the associated eigenvector

$$\left( \begin{array}{c}2-i \end{array} \right)$$$$\left( \begin{array}{c}1 \\2-i \end{array} \right)$$

and got

$$\left( \begin{array}{c}2+i \\5 \end{array} \right)$$

So not sure whats going on

6. Sep 22, 2009

### boneill3

I have had a look at the solution given and this is the way that was shown.

from

$$\left( \begin{array}{cc}4 & -5 \\1 & 0 \end{array} \right)$$

you try and diagonalise the matrix.

So

$$\left( \begin{array}{cc}4 & -5 \\1 & 0 \end{array} \right)$$

becomes

$$\left( \begin{array}{cc}0 & 0 \\1 & 0 \end{array} \right)$$

and you solve for zero with the eigen values obtained earlier to produce the eigenvectors using

$$\left( \begin{array}{c}1 & 0-\lambda \end{array} \right)$$

so solve

$x-(2-i)y=0$

you do the same with the other eigenvalue $2+i$

I don't know how they came to the following eigen vector did they just pick the numbers or was there a particular reason. They gave the eigenvector as

$$\left( \begin{array}{cc}5 \\ 2-i\end{array} \right)$$
and
$$\left( \begin{array}{cc}5 \\ 2+i\end{array} \right)$$

I suppose there are infinite solutions but maybe because of the -5 above?

They use the vectors and call this the transition matrix

$$\left( \begin{array}{cc}5 & 5 \\2-i & 2+i \end{array} \right)$$

Next they take the inverse of the transition matrix

$$\left( \begin{array}{cc}\frac{1}{10}-\frac{1}{5}\times 1 & \frac{1}{2}\times i \\\frac{1}{10}+\frac{1}{5}\times 1 & -\frac{1}{2}\times i \end{array} \right)$$

and multiply that by the original matrix then the transition matrix.

$$\left( \begin{array}{cc}4 & -5 \\1 & 0 \end{array} \right)$$

and

$$\left( \begin{array}{cc}5 & 5 \\2-i & 2+i \end{array} \right)$$

which gives the diagonalised matrix

$$\left( \begin{array}{cc}2+i & 0 \\0 & 2-i \end{array} \right)$$

which proves that they are the right eigenvectors.

Can someone please let me know why the chose those particular eigienvectros ?
As I could not simulate the last with those vectors.

regards