Yont's question at Yahoo Answers (eigenvector)

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In summary, the matrix A = [{-7,3},{0,-4}] has eigenvalues -4 and -7. The associated eigenvectors with -7 are vectors of the form $\begin{bmatrix} x \\ 0 \end{bmatrix}$ with $x \neq 0$.
  • #1
Fernando Revilla
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Here is the question:

The matrix A = [{-7,3},{0,-4}] has eigenvalues -4 and -7. What are the associated eigenvectors with -7. I figured out that -4 is with (1,1) but i can't figure out -7

Here is a link to the question:

Find the eigenvector? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
$\begin{bmatrix} -7 & 3 \\ 0 & -4 \end{bmatrix} * x = \begin{bmatrix}7x \\ 7 y\end{bmatrix}$

or

-7x + 3y = -7x or 3y = 0, or y = 0so vectors of the form , $\begin{bmatrix} x \\ 0 \end{bmatrix} $
with $x \not = 0 $
(Sorry fernando if i hijacked this, i know how much you love answering these but I've had 6 cups of coffee and i need to do something from going crazy)
 
  • #3
jakncoke said:
$\begin{bmatrix} x \\ 0 \end{bmatrix} $

Certainly, by definition, $ \begin{bmatrix}{x}\\y\end{bmatrix}\neq \begin{bmatrix}{0}\\0\end{bmatrix}$ is an eigenvector of $A$ associated to $\lambda=-7$ if and only if:

$\begin{bmatrix}{-7}&{\;\;3}\\{\;\;0}&{-4}\end{bmatrix}\begin{bmatrix}{x}\\y\end{bmatrix}=(-7)\begin{bmatrix}{x}\\y\end{bmatrix}$

You'll easily get the system $\{y=0$ so, all the eigenvalues associated to $\lambda=-7$ are

$\begin{bmatrix}{\alpha}\\0\end{bmatrix}$ with $\alpha\neq 0$.
 
  • #4
Fernando Revilla said:
Certainly, by definition, $ \begin{bmatrix}{x}\\y\end{bmatrix}\neq \begin{bmatrix}{0}\\0\end{bmatrix}$ is an eigenvector of $A$ associated to $\lambda=-7$ if and only if:
$\begin{bmatrix}{-7}&{\;\;3}\\{\;\;0}&{-4}\end{bmatrix}\begin{bmatrix}{x}\\y\end{bmatrix}=(-7)\begin{bmatrix}{x}\\y\end{bmatrix}$.

You'll easily get the system $\{y=0$ so, all the eigenvalues associated to $\lambda=-7$ are

$\begin{bmatrix}{\alpha}\\0\end{bmatrix}$ with $\alpha\neq 0$.

yes sir, i fixed it. you are absolutely right.
 
  • #5


Hello Yont,

Thank you for your question about eigenvectors. As you correctly pointed out, the eigenvalue -4 is associated with the eigenvector (1,1). To find the eigenvector associated with -7, we can use the following formula:

Av = λv

Where A is the matrix, v is the eigenvector, and λ is the eigenvalue. In this case, we have:

[-7,3][x,y] = -7[x,y]

Simplifying this equation, we get:

-7x + 3y = -7x

3y = 0

This means that y can be any value, as long as x is 0. Therefore, the eigenvector associated with -7 is [0,1].

I hope this helps answer your question. Keep up the good work in exploring eigenvectors and their properties!

Best,
[Your name]
 

Related to Yont's question at Yahoo Answers (eigenvector)

1. What is an eigenvector?

An eigenvector is a vector that does not change direction when a linear transformation is applied to it. In other words, it is a special vector that is only scaled by the transformation and not rotated.

2. What is the importance of eigenvectors in science and mathematics?

Eigenvectors are important in many areas of science and mathematics, such as linear algebra, quantum mechanics, and data analysis. They are used to understand the behavior of complex systems and to find solutions to various problems.

3. How do you calculate eigenvectors?

To calculate eigenvectors, you need to first find the eigenvalues of a matrix. Then, for each eigenvalue, you can solve a system of equations to find the corresponding eigenvector.

4. Can eigenvectors have negative values?

Yes, eigenvectors can have negative values. The sign of an eigenvector does not affect its properties or calculations. However, in some cases, the negative sign may indicate a change in direction.

5. What is the relationship between eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are closely related. An eigenvector is associated with a specific eigenvalue, and the eigenvalue represents the scaling factor for the eigenvector. In other words, the eigenvalue determines how the eigenvector is transformed by a matrix.

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