# Trouble with complex eigenvector

1. Nov 21, 2006

### Mindscrape

The problem is to solve the differential equation where
$$\mathbf{x'} = \left( \begin{array}{cc} 1 & -5\\ 1 & -3 \end{array} \right) \mathbf{x}$$
given that
$$\mathbf{x(0)} = \left( \begin{array}{cc} 5 \\ 4 \end{array} \right)$$

The eigenvalues are easy to find, and they are:
$$\lambda = 1 + i$$
and
$$\lambda = 1 - i$$

I am having trouble reducing the complex matrix with Gauss-Jordan elimination, and was wondering if I could get some pointers. Here is what I have done:
By pluging in the first eigenvalue the matrix to reduce is

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

then I figure that I should multiply by i on the top and divide by -4-i on the bottom so that the pivots are 1 and real:
This makes
$$\left( \begin{array}{cc} 1 & -5i\\ \frac{1}{-4-i} & 1 \end{array} \right)$$

I tried to take it from here, but I wasn't able to get anywhere. I tried complex conjugates in various places, and multiplying i here and there, but no luck. Most of the difficulty I am having is just reducing the second row to the zero row. I can get a zero at (2,1), but getting (2,2) to zero is proving difficult.

Last edited: Nov 21, 2006
2. Nov 22, 2006

### Mindscrape

Okay, here is the rest of my work in the hope that someone can point out where I went wrong:

Take the matrix with the first eigenvalue plugged in, which is

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

now multiply the top row by i

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

multiply the bottom row by 5

$$\left( \begin{array}{cc} 1 & -5i\\ 5 & -20 - 5i \end{array} \right)$$

in row 2, subtract row 1

$$\left( \begin{array}{cc} 1 & -5i\\ 4 & -20 \end{array} \right)$$

in (1/4)row 2, subtract row 1

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

multiply the top row by (1-i) and the bottom by -1

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

then this gives

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

It is almost Gauss Jordan, but the second pivot is complex, and this isn't even the result I want. I want the bottom row to be zeros so that I can compose the eigenvector.

3. Nov 22, 2006

### rocketturtle

hmm, it looks like your eigenvalues may be slightly off

4. Nov 23, 2006

### HallsofIvy

Staff Emeritus
Yes, the eigenvalues are easy to find- but they are NOT 1+ i and 1- i!

Last edited: Nov 24, 2006
5. Nov 23, 2006

### Benny

I can't remember much linear algebra so I can't justify the following. But anyway, you don't need to row reduce. Simply solve either one of the equations and the solution will automatically satisfy the other equation. In other words, you only need to work with one equation to extract the eigenvector.

6. Nov 26, 2006

### Mindscrape

Whoops, just realized I copied the values down wrong after I solved for them. They should be:
$$\lambda = -1 + i$$
and
$$\lambda = -1 - i$$

Now I can solve.