Question regarding eigenvectors

[arhzz]

Homework Statement

Eigenvectors

Relevant Equations

-

So I have been studying for my upcoming math exam and a lot of the problems require to find eigenvalues/eigenvectors.Now the question I have is the following;

Take a look at this matrix

$$ \left[ \begin{matrix} 6 & -3 \\\ 3 & -4 \end{matrix} \right] $$

Now the eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = -3$

Now to find the eigenvectors I did this.

First I subtracted 5 from the diagonal of the original matrix and it looks like this.

$$ \left[ \begin{matrix} 1 & -3 \\\ 3 & -9 \end{matrix} \right] $$

Now to find the eigenvectors this should be the way

$$ \left[ \begin{matrix} 1 & -3 \\\ 3 & -9 \end{matrix} \right] \cdot \begin{pmatrix} x_1 \\\ x_2 \end{pmatrix} =\begin{pmatrix} 0 \\\ 0 \end{pmatrix} $$

And the equations look like this

$x_1 - 3x_2 = 0$
$3x_1 - 9x_2 = 0$

If we multiply the first equation with 3 and subract that from the second equation; we get

$x_1 - 3x_2 = 0$
$0 0 = 0$

Now we can say that $x_1 = 3x_2$ and let x1 = 1 we get that the eigenvector is

$v_1 = (1,3)$

Now the solutions say it should be $v_1 = (3,1)$ the same simply rearanged.My question is are both eigenvectors valid? Does it matter if its (1,3) or (3,1) because I'm susposed to build a fundamental matrix and I'm not sure if this plays a role.

 

[fresh_42]

An eigenvector satisfies the equation $A\cdot \vec{v} = \lambda \cdot \vec{v}$ or $(A-\lambda \cdot I)\cdot \vec{v}=0$ where $I$ is the identity matrix and $\lambda$ the variable, the eigenvalue we want to know.

Now you should solve this equation system. If you already had the determinant then use it here.

 

[Mark44]

Homework Statement:: Eigenvectors
Relevant Equations:: -

So I have been studying for my upcoming math exam and a lot of the problems require to find eigenvalues/eigenvectors.Now the question I have is the following;

Take a look at this matrix

$$ \left[ \begin{matrix} 6 & -3 \\\ 3 & -4 \end{matrix} \right] $$

Now the eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = -3$

Now to find the eigenvectors I did this.

First I subtracted 5 from the diagonal of the original matrix and it looks like this.

$$ \left[ \begin{matrix} 1 & -3 \\\ 3 & -9 \end{matrix} \right] $$

Now to find the eigenvectors this should be the way

$$ \left[ \begin{matrix} 1 & -3 \\\ 3 & -9 \end{matrix} \right] \cdot \begin{pmatrix} x_1 \\\ x_2 \end{pmatrix} =\begin{pmatrix} 0 \\\ 0 \end{pmatrix} $$

And the equations look like this

$x_1 - 3x_2 = 0$
$3x_1 - 9x_2 = 0$

If we multiply the first equation with 3 and subract that from the second equation; we get

$x_1 - 3x_2 = 0$
$0 0 = 0$

Now we can say that $x_1 = 3x_2$ and let x1 = 1 we get that the eigenvector is

$v_1 = (1,3)$

No. If $x_1 = 1$ then $x_2 = 1/3$.

It's easier to set $x_2 = 1$ which gives $x_1 = 3$.

Now the solutions say it should be $v_1 = (3,1)$ the same simply rearanged.My question is are both eigenvectors valid? Does it matter if its (1,3) or (3,1) because I'm susposed to build a fundamental matrix and I'm not sure if this plays a role.

<3, 1> is an eigenvector; <1, 3> is not.

 

[Orodruin]

Apart from what has already been said, when you construct your equations they will be linearly dependent by construction. This means you can just throw one of them away in the case of a 2x2 matrix - unless the equation is 0 = 0 - then use the other (although be careful in the case of more than 2x2 matrices not to throw away equations that are linearly independent from the rest ...) There is no need for 2x2 matrices to go around eliminating by multiplying and subtracting.

 

[arhzz]

Uh your right,I did my algebra wrong. But also good to know that I can just focus on one equation. Thanks for the help!

 

[WWGD]

Homework Statement:: Eigenvectors
Relevant Equations:: -

So I have been studying for my upcoming math exam and a lot of the problems require to find eigenvalues/eigenvectors.Now the question I have is the following;

Take a look at this matrix

$$ \left[ \begin{matrix} 6 & -3 \\\ 3 & -4 \end{matrix} \right] $$

Now the eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = -3$

Now to find the eigenvectors I did this.

First I subtracted 5 from the diagonal of the original matrix and it looks like this.

$$ \left[ \begin{matrix} 1 & -3 \\\ 3 & -9 \end{matrix} \right] $$

Now to find the eigenvectors this should be the way

$$ \left[ \begin{matrix} 1 & -3 \\\ 3 & -9 \end{matrix} \right] \cdot \begin{pmatrix} x_1 \\\ x_2 \end{pmatrix} =\begin{pmatrix} 0 \\\ 0 \end{pmatrix} $$

And the equations look like this

$x_1 - 3x_2 = 0$
$3x_1 - 9x_2 = 0$

If we multiply the first equation with 3 and subract that from the second equation; we get

$x_1 - 3x_2 = 0$
$0 0 = 0$

Now we can say that $x_1 = 3x_2$ and let x1 = 1 we get that the eigenvector is

$v_1 = (1,3)$

Now the solutions say it should be $v_1 = (3,1)$ the same simply rearanged.My question is are both eigenvectors valid? Does it matter if its (1,3) or (3,1) because I'm susposed to build a fundamental matrix and I'm not sure if this plays a role.

You can always check, with your matrix. Evaluate it at $(3,1)$. Is $(3,1)$ an Eigenvector associated to$5$? Is $A[3,1]^{T}=5*(3,1)^{T}$ ?

Note: $(3,1)^{T}$ is just the vector $(3,1)$, written as a column vector, instead of as a row vector, to make the requirements that $Av$ ; for $A$ being $m \times n$ , and the vector being $n \times 1$.

 

[DaveE]

Uh your right,I did my algebra wrong. But also good to know that I can just focus on one equation. Thanks for the help!

You can also easily check your own work once you think you have an answer. Does Av=λv? It's not hard in two dimensions.

This is how the real world works once you graduate. There is no answer key when you're building a bridge. Your boss doesn't want to have to do all of the same work you do to see if you've made mistakes. All of us make mistakes from time to time, it's what humans do. So engineering solutions should always be verified whenever possible. You will find that the amount of effort expended to check your work is much less than the effort required when mistakes aren't noticed until later.

 

[arhzz]

Your right of course (as per usual).Checking the answer is really simple,I've been able to verify that (1,3) is not an eigenvector. Thanks for the help as always,cheers!