Can you help me find Eigenvectors for this matrix?

  • Thread starter M55ikael
  • Start date
In summary, the conversation is about a person struggling with solving an eigenvector equation for a matrix and finding the eigenvectors. The matrix is simplified and the resulting equations are presented. The solution is then shown to be a simple calculation error and the correct eigenvectors are given.
  • #1
M55ikael
17
0

Homework Statement



Hi! I have this frustrating equation I can't solve even though I have the answer.
I'm trying to make two Eigenvectors for my matrix. I have found λ1 and λ2, but I can't solve the resulting equations.

Homework Equations



[itex]A = \begin{bmatrix} 2 & 1 \\ -\frac{1}{2} & 1 \end{bmatrix}[/itex]


λ1= 3/2 + j/2 , λ2= 3/2 - j/2

[itex]\begin{bmatrix} 2 -(3/2 + j/2) & 1 \\ -\frac{1}{2} & 1 - (3/2 + j/2) \end{bmatrix}[/itex] * E1 = 0



Attempt at solution:

E1
α(-1-j) + 2β = 0
-α + β(-1-j) = 0


E2
α(-1+j) + 2β = 0
-α + β(-1+j) = 0

I can't make any sense out of those equations though.

Answers should be
[itex]E1 = \begin{bmatrix} 2 \\ -1+j \end{bmatrix}[/itex] * c

[itex]E2 = \begin{bmatrix} 2 \\ -1-j \end{bmatrix}[/itex] * t

c and t are different from 0
 
Last edited:
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  • #2
M55ikael said:

Homework Statement



Hi! I have this frustrating equation I can't solve even though I have the answer.
I'm trying to make two Eigenvectors for my matrix. I have found λ1 and λ2, but I can't solve the resulting equations.

Homework Equations



[itex]A = \begin{bmatrix} 2 & 1 \\ -\frac{1}{2} & 1 \end{bmatrix}[/itex]


λ1= 3/2 + j/2 , λ2= 3/2 - j/2

[itex]\begin{bmatrix} 2 -(3/2 + j/2) & 1 \\ -\frac{1}{2} & 1 - (3/2 + j/2) \end{bmatrix}[/itex] * E1 = 0
This looks fine. Maybe you got fouled up in the arithmetic. The matrix can be simplified a bit, which will make the subsequent work easier.

[itex]\begin{bmatrix} 1/2 - i/2 & 1 \\ -1/2 & -1/2 - i/2 \end{bmatrix}[/itex]

I'm using i for the imaginary unit...

This row reduces to
[itex]\begin{bmatrix} 1 & 1 + i \\ 0 & 0 \end{bmatrix}[/itex]

The top row says that x1 = -(1 + i)x2, so an eigenvector is
[itex]\begin{bmatrix} -1 - i \\ 1 \end{bmatrix}[/itex]


M55ikael said:
Attempt at solution:

E1
α(-1-j) + 2β = 0
-α + β(-1-j) = 0


E2
α(-1+j) + 2β = 0
-α + β(-1+j) = 0

I can't make any sense out of those equations though.

Answers should be
[itex]E1 = \begin{bmatrix} 2 \\ -1+j \end{bmatrix}[/itex] * c

[itex]E2 = \begin{bmatrix} 2 \\ -1-j \end{bmatrix}[/itex] * t

c and t are different from 0
 
  • #3
Thanks, I allways knew it was pretty simple. But it's so easy to get your pluses and minuses mixed up, I couldn't get it reduced properly.
 

1. What is an equation?

An equation is a mathematical statement that shows the relationship between two or more quantities. It consists of variables, constants, and mathematical operations.

2. How do I solve an equation?

To solve an equation, you need to isolate the variable on one side of the equation by performing the same mathematical operations on both sides. This will result in finding the value of the variable.

3. Can I solve any equation?

Yes, you can solve any equation as long as it follows the rules of algebra. However, some equations may be more complex and require advanced techniques to solve.

4. What if I get a negative answer when solving an equation?

A negative answer in an equation may mean that you made a mistake during the solving process. Double-check your work and make sure you followed the correct steps. If the negative answer is expected, it may indicate that the solution is a negative number.

5. How can I check if my solution is correct?

You can check your solution by plugging it back into the original equation and solving it. If the result is a true statement, then your solution is correct. Another way to check is by using a graphing calculator to plot the equation and see if the solution aligns with the intersection point on the graph.

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