How do I find the eigenvectors for this matrix?

In summary, we can find the eigenvalues of a matrix by solving the equation Av = \lambda v, where A is the matrix, v is the eigenvector, and \lambda is the eigenvalue. To find the eigenvectors, we can plug in the eigenvalues into the equation (A-I\lambda)v = 0 and solve for v. In the given examples, the eigenvalues of the matrix \left(\begin{array}{cc}0 & \frac{1}{2}\\ \frac{1}{2} & 0\end{array}\right) are \lambda_1 = \frac{1}{2} and \lambda_2 = -\frac{1}{
  • #1
mprm86
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The eigenvalues of the matrix [tex] \left(\begin{array}{cc}0 & \frac{1}{2}\\ \frac{1}{2} & 0\end{array}\right) [/tex] are [itex] \lambda_1 = \frac{1}{2} [/itex] and [itex] \lambda_2 = -\frac{1}{2} [/itex]
The problem here is that I have no idea of how to calculate the eigenvectors. Could some one please explain me, in detail, how do I find the eigenvectors?
Thanks in advance. (sorry for posting the thread twice)
 
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  • #2
Those aren't the eigenvalues...

You should get [itex]\lambda_1 = 0, \; \lambda_2 = 1/2[/itex].

Edit: Okay, now that you've fixed your post, solve the equations [itex]Av = \lambda_1 v[/itex] and [itex]Av = \lambda_2 v[/itex].
 
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  • #3
mprm86 said:
The eigenvalues of the matrix [tex] \left(\begin{array}{cc}0 & \frac{1}{2}\\ \frac{1}{2} & 0\end{array}\right) [/tex] are [itex] \lambda_1 = \frac{1}{2} [/itex] and [itex] \lambda_2 = -\frac{1}{2} [/itex]
The problem here is that I have no idea of how to calculate the eigenvectors. Could some one please explain me, in detail, how do I find the eigenvectors?
Thanks in advance. (sorry for posting the thread twice)
There is no zero eigenvalue, as the transformation of this matrix is 1-1 on R2. Specifically, the transformation includes an inversion of one axis, as it maps a scaling of the positive x-axis to the positive y-axis and vice-versa for the positive y-axis (Remember that the columns of the matrix form a basis for the image of the transformation). Ignoring the scaling factor of 1/2, it is then easy to see that two distinct eigenvectors are (1,1) (rotate the positive x-axis into the positive y-axis, then invert the old y-axis) and (1,-1) by the same visualization is mapped to its negative. Noting that the scaling factor is 1/2, we then find the eigenvalues are 1/2 and -1/2 as you found.
 
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  • #4
In the case that the transformation of the matrix is not very nice, you will want an algebraic way of getting eigenvectors. This can be done by simply plugging each eigenvalue you get back into the equation [tex](A - I\lambda)v = 0[/tex] where A is the matrix, lambda is the eigenvalue, I is the identity matrix, and v is an eigenvector for lambda and A, and solve for v. For example, when [tex]\lambda = \frac{1}{2}[/tex], you will want to solve the equation:
[tex]\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{array}\right) \left(\begin{array}{c}v^1\\ v^2\end{array}\right) = 0[/tex]
Note that you are solving for a one-dimensional subspace (not just a single vector), so you will have one free variable.
 
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  • #5
This is where I get stuck. If I solve the equation [tex]\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{array}\right) \left(\begin{array}{c}v^1\\ v^2\end{array}\right) = 0[/tex], I get the next system of equations:
[tex] -1/2v^1 + 1/2v^2 = 0 ... (1) [/tex]
[tex] 1/2v^1 - 1/2v^2 = 0 ... (2) [/tex]
and I think the solution for this system is [tex] v^1 = v^2 [/tex], so I don´t know how to interpret this.
 
  • #6
correct. It means that your eigenvector is of the form

[tex](v_1 \ , v_1)^T = v_1(1 \ , \ 1)^T[/tex]

ie. a basis for the eigenvectors corresponding to the eigenvalue [itex] 1 /2 [/itex] is

[tex](1\ ,\ 1)^T.[/tex]
 
  • #7
So, this means that the first eigenvector will be (1/2, 1/2)?
 
  • #8
that is fine as well. Using [itex](1 \ , \ 1), \ ( 1/2\ ,\ 1/2), [/itex] or [itex](1001412\ , \ 1001412)[/itex] are all equivalent.
 
  • #9
Ok, i got that. Just help me with this other example. The eigenvalues of the matrix [tex] \left(\begin{array}{cc}5 & -3\\ -3 & 5\end{array}\right) [/tex] are [itex] \lambda_1 = 2 [/itex] and [itex] \lambda_1 = 8 [/itex], and, according to my book, the eigenvectors are [itex] (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) [/itex] and [itex] (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) [/itex]. I don't get how did they got them.
 
  • #10
Well, what did you get when you tried to find them? What did you do to find your answers?
 

FAQ: How do I find the eigenvectors for this matrix?

1. What are eigenvectors and why are they important?

Eigenvectors are a type of vector that represent a special set of directions in a vector space. They are important because they have a wide range of applications in fields such as physics, computer science, and engineering. They allow us to understand and solve complex problems involving systems of linear equations.

2. How do you find eigenvectors?

To find eigenvectors, we first need to find the eigenvalues of a given matrix. This can be done by solving the characteristic equation of the matrix. Once we have the eigenvalues, we can plug them back into the original matrix to find the corresponding eigenvectors. This process can be done by hand or using specialized software.

3. What is the significance of the eigenvector-eigenvalue relationship?

The eigenvector-eigenvalue relationship is significant because it allows us to decompose a complex matrix into simpler components. Eigenvectors represent the directions along which a matrix only stretches or shrinks, while eigenvalues represent the amount of stretching or shrinking. This makes it easier to analyze and understand the behavior of a matrix.

4. Can eigenvectors be negative?

Yes, eigenvectors can be negative. The magnitude and direction of an eigenvector are what matter, not the sign. In fact, negative eigenvectors can be useful in certain applications, such as in quantum mechanics.

5. What are some real-world applications of eigenvectors?

Eigenvectors have numerous real-world applications, such as in image and signal processing, data analysis, and network analysis. They are also used in machine learning algorithms, computer graphics, and quantum mechanics. In physics, eigenvectors are used to describe the behavior of systems in quantum mechanics and in analyzing the stability of physical systems such as bridges and buildings.

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