MHB Yont's question at Yahoo Answers (eigenvector)

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The matrix A = [{-7,3},{0,-4}] has eigenvalues -4 and -7, with the eigenvector associated with -4 identified as (1,1). For the eigenvalue -7, the associated eigenvectors take the form of vectors where y = 0, resulting in eigenvectors of the type [{x}, 0] with x not equal to 0. This means any non-zero value for x will yield a valid eigenvector for -7. The discussion confirms the solution and clarifies the conditions for eigenvectors related to the eigenvalue -7.
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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$\begin{bmatrix} -7 & 3 \\ 0 & -4 \end{bmatrix} * x = \begin{bmatrix}7x \\ 7 y\end{bmatrix}$

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)
 
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 $\alpha\neq 0$.
 
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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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