MHB Yont's question at Yahoo Answers (eigenvector)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Eigenvector
AI Thread Summary
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
Gold Member
MHB
Messages
631
Reaction score
0
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.
 
Mathematics news on Phys.org
$\begin{bmatrix} -7 & 3 \\ 0 & -4 \end{bmatrix} * x = \begin{bmatrix}7x \\ 7 y\end{bmatrix}$

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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top