# Trouble with complex eigenvector

• Mindscrape
In summary, you are having difficulty reducing the complex matrix, and were wondering if you could get some pointers. You tried reducing the matrix by plugging in the first eigenvalue, but then you figured you should multiply by i on the top and divide by -4-i on the bottom, which made the pivots 1 and real. This made it almost Gauss-Jordan elimination, but the second pivot is complex, and this isn't even the result you want. You can solve for the eigenvector either by solving one equation or by solving the two equations and using the solution to the other equation.
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:
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.

hmm, it looks like your eigenvalues may be slightly off

Mindscrape said:
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
$$\lambda = 1 - i$$

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

Last edited by a moderator:
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.

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.

## 1. What is an eigenvector and why is it important?

An eigenvector is a vector that does not change its direction when multiplied by a particular matrix. It is important because it represents the directions along which a transformation acts by simply stretching or compressing, without changing the direction of the vector.

## 2. How do I know if a matrix has complex eigenvectors?

A matrix has complex eigenvectors if it has complex eigenvalues. This can be determined by finding the characteristic polynomial of the matrix and solving for its roots. If any of the roots are complex numbers, then the matrix has complex eigenvectors.

## 3. What causes trouble with complex eigenvectors?

The main cause of trouble with complex eigenvectors is the difficulty in visualizing and understanding complex numbers. Complex numbers have both a real and imaginary component, which can make it challenging to interpret the direction and magnitude of the eigenvectors.

## 4. How can I work with complex eigenvectors in practical applications?

In practical applications, complex eigenvectors can be used to represent oscillating or rotating systems. They are also useful in solving differential equations and in fields such as quantum mechanics and signal processing.

## 5. Are there any techniques for simplifying complex eigenvectors?

Yes, there are techniques such as taking the real and imaginary parts of the complex eigenvector to create a real-valued vector, or using geometric visualization tools such as Argand diagrams to better understand the complex eigenvector's direction and magnitude.

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