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Carissa15
 one year ago
I have a few questions about eigenvalues and eigenvectors in matrices.
Carissa15
 one year ago
I have a few questions about eigenvalues and eigenvectors in matrices.

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Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Matrix b} \[\left[\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right]\]

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Matrix a)\[\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3how far did you get with the eigenvalues of \[\left[\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right]\]

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I really haven't done anything with eigenvectors or eigenvalues. Very new to me. Seems though if I

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3are you familiar with the idea that you need to find the determinant of the matrix \(\LARGE A  \lambda I\) to get the eigenvalues?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0\[\det( \lambda \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right])=0\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3it should be the other way around A  (lambda)*I not (lambda)*I  A

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so you need to solve \[\Large \det( \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]\lambda\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right])=0\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 Yes, it is. I had 2 profs who used different ways to find out the eigenvalues. One used lambda A and other used A  lambda. Both worked well.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3oh I did not know that

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3well however you solve the equation, what values of lambda do you get?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0just working through now

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I get \[\lambda^24\lambda45=0\] not sure where to go from there

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3use the quadratic formula to solve for lambda

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3you can also factor the left side if you want

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0is the quadratic formula \[\lambda^2\lambda4=0\]?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3if you factored, you'll find that \[\Large \lambda^24\lambda45=0\] turns into \[\Large (\lambda9)(\lambda+5)=0\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so you agree that \[\Large \lambda = 9 \text{ or } \lambda = 5\] or no?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0so these are the eigenvalues?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I am not sure how to now find eigenectors using these

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large A\vec{v} = \lambda \vec{v}\] \[\Large A\vec{v}  \lambda \vec{v}=0\] \[\Large A\vec{v}  \lambda I\vec{v}=0\] \[\Large (A  \lambda I)\vec{v}=0\] That last equation leads us to this matrix equation \[\Large \begin{bmatrix}2\lambda & 7\\7&2\lambda\end{bmatrix} \vec{v} = \begin{bmatrix}0\\0\end{bmatrix}\] which leads to this augmented matrix \[\Large \left[\begin{array}{ccc}2\lambda & 7 & 0 \\7 & 2\lambda & 0\end{array}\right]\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3If lambda = 9, then \[\Large \left[\begin{array}{ccc}2\lambda & 7 & 0 \\7 & 2\lambda & 0\end{array}\right]\] \[\Large \left[\begin{array}{ccc}29 & 7 & 0 \\7 & 29 & 0\end{array}\right]\] \[\Large \left[\begin{array}{ccc}7 & 7 & 0 \\7 & 7 & 0\end{array}\right]\] row reduce the augmented matrix until it's in rref form

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0So we would end up with R1+R2 and all values zero?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3you do compute R1+R2, yes but we only need to do that for one row R1+R2 = [0 0] replace R2 with that result \[\Large \left[\begin{array}{ccc}7 & 7 & 0 \\7&7 & 0\end{array}\right]\] \[\Large \left[\begin{array}{ccc}7 & 7 & 0 \\0&0 & 0\end{array}\right]\begin{array} \ \\ R1+R2 \rightarrow R2\end{array}\]

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Oh yes, of course. not both oops.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3the top row 7 7 0 means 7x+7y = 0 solve for y to get 7y = 7x y = x so the solution set is the column vector \(\LARGE \left[\begin{array}{c}x \\x\end{array}\right]\) this describes all of the eigenvectors for the eigenvalue lambda = 9

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3a lot of books may say let x = 1, so a particular eigenvector for lambda = 9 is \[\LARGE \left[\begin{array}{c}1 \\1\end{array}\right]\] all other eigenvectors for lambda = 9 are simply scalar multiples of that column vector

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Would this always be the case? Or do you sometimes get actual values instead of x? And Do I then follow the same steps for lambda 5?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yes I've had teachers who would assign values to x. 1 is the smallest possible to allow for other multiples to be generated

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3and yes you follow the same steps for lambda = 5

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0ok great. Thank you. I will work my way through them and see how I go. :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3ok tell me what you get

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0for lambda 5 i got\[\left[\begin{matrix}7 & 7 \\ 7 & 7\end{matrix}\right]\\] with Identity matrix

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0then did R1R2 gives new R2 of all zeros.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I then got 7x7y=0, which then ends up with the same result as lambda 9

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3I'm getting the same, so you have 7 7 0 up top over 0 0 0 the top row implies that 7x + 7y = 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.37x7y=0 is incorrect

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0it is always x+y=0 ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3I don't know what you mean by "always", but 7x + 7y = 0 does reduce to x+y = 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3then x+y = 0 turns into y = x when you solve for y

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so \[\Large \left[\begin{array}{ccc}x \\ y\end{array}\right]=\left[\begin{array}{ccc}x \\ x\end{array}\right]\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3if we let x = 1, then \(\Large \left[\begin{array}{ccc}1 \\ 1\end{array}\right]\), and all of its scalar multiples, represent the other set of eigenvectors.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3I recommend computing \[\Large A*\vec{v} = \lambda*\vec{v}\] \[\Large \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]*\left[\begin{matrix}1 \\ 1\end{matrix}\right] = 9*\left[\begin{matrix}1 \\ 1\end{matrix}\right] \] to verify the first eigenvalue + eigenvector (lambda = 9)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3same for lambda = 5

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I have taken the same steps for matrix b) but get stuck as I cannot factor to find eigenvalues. Please help.

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right]\] I end up with \[\lambda^22\lambda1=0\] but get stuck trying to factor to find eigenvalues

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0eigenvalues : \(\lambda = 1\pm \sqrt{2}\) eigenvectors : \[\begin{bmatrix} 3(1+\sqrt{2})&2\\1&1(1+\sqrt{2})\end{bmatrix} \begin{pmatrix} x_1\\x_2\end{pmatrix}= 0\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0then they are same, just pick one row to calculate \((2\sqrt 2 )x_1= 2 x_2\) so if \(x_1 = 2 , then~~x_2 = 2\sqrt2\) that is \(\left[\begin{matrix} 2\\2\sqrt 2\end{matrix}\right]\) is your eigenvector w.r.t \(1+\sqrt2\)

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0Thank you everyone. Solved :)
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