Find the eigenvalues and eigenvectors

In summary, you found the eigenvalues and eigenvectors for the matrix A using the characteristic polynomial equation. You found that the first eigenvalue is 1 and the second eigenvalue is −1. You found that the eigenvector associated with the first eigenvalue is (1, −i). You found that the eigenvector associated with the second eigenvalue is (1, −i).
  • #1
Mutatis
42
0

Homework Statement



Find the eigenvalues and eigenvectors fro the matrix: $$
A=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$.

Homework Equations



Characteristic polynomial: ## \nabla \left( t \right) = t^2 - tr\left( A \right)t + \left| A \right|## .

The Attempt at a Solution



I've found the eigenvalues doing through the characteristic polynomial equation above: $$ \lambda_1 = 1 $$ $$ \lambda_2 = -1 $$.
Then, to get the eigenvector associated to ## \lambda_1## the equation ##M v_1 = 0## must be satisfied, $$ \begin{pmatrix} \left(0-1\right) & -i \\ i & \left(0-1\right) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0$$.
It leads me to a system that I'm having trouble to solve it: $$ \begin{cases} -x-iy=0 \\ ix-y=0 \end{cases} $$.
I don't know what to do next, please help me!
 
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  • #2
Why are you having trouble ? What are you expecting to find ?
 
  • #3
As I'm sure you realized, these equations are linearly dependent so the system has infinitely many solutions. So parametrize it, let ##y## be anything. Say ##y = a##. Solve for ##x##. That's a valid eigenvector. Any multiple of that is a valid eigenvector.
 
  • #4
RPinPA said:
parametrize it, let yyy be anything. Say y=ay=ay = a. Solve for xxx. That's a valid eigenvector. Any multiple of that is a

I did what you've said ## y=1## then ##x=1/i=-i##, so I got ## v_1=(1, -i)##. When I put this vector in the matrix to verify ##Mv_1=0## it leads me to a non-zero value...
 
  • #5
Mutatis said:
I did what you've said ## y=1## then ##x=1/i=-i##, so I got ## v_1=(1, -i)##. When I put this vector in the matrix to verify ##Mv_1=0## it leads me to a non-zero value...

That's odd: when I put ##x = -i## and ##y = 1## into ##-x - iy## I get ##0##, exactly as wanted. The second left-hand-side is just ##i \times## the first left-hand-side, so it will equal ##0## also.
 
  • #6
Mutatis said:
I did what you've said ##y=1## then ##x=1/i=−i##, so I got ##v_1=(1,−i)##.

Are you sure? If ##x## is ##-i## and ##y## is ##1##, you would write that as ##(1, -i)##?
 
  • #7
Oh! Guys I'm sorry I wrote the values wrong! Now I understand what I was doing wrong. Thank you very much guys!
 

FAQ: Find the eigenvalues and eigenvectors

1. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are mathematical concepts used in linear algebra to describe the behavior of a linear transformation. Eigenvalues are the values that, when multiplied by a vector, produce a scalar multiple of that vector. Eigenvectors are the corresponding vectors that are scaled by the eigenvalues.

2. Why is it important to find eigenvalues and eigenvectors?

Finding eigenvalues and eigenvectors is important because they provide valuable information about the behavior of a linear transformation. They can be used to determine the stability of a system, identify important features of a dataset, and simplify complex calculations in linear algebra.

3. How do you find eigenvalues and eigenvectors?

To find eigenvalues and eigenvectors, you first need to set up a matrix representing the linear transformation. Then, you can use various methods such as the characteristic polynomial or Gaussian elimination to find the eigenvalues. Once the eigenvalues are known, the corresponding eigenvectors can be found by solving a system of linear equations.

4. What is the relationship between eigenvalues and eigenvectors?

The relationship between eigenvalues and eigenvectors is that each eigenvalue corresponds to a specific eigenvector. The eigenvector is scaled by the eigenvalue when it undergoes the linear transformation represented by the matrix. This means that the eigenvector is only changed in magnitude, not in direction, when multiplied by the eigenvalue.

5. Can a matrix have more than one set of eigenvalues and eigenvectors?

Yes, a matrix can have multiple sets of eigenvalues and eigenvectors. This occurs when the matrix has repeated eigenvalues, meaning that there are multiple eigenvectors corresponding to the same eigenvalue. In this case, the eigenvectors form a basis for the eigenspace associated with that eigenvalue.

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