Carissa15
  • Carissa15
I have a few questions about eigenvalues and eigenvectors in matrices.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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Carissa15
  • Carissa15
Matrix b} \[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\]
Carissa15
  • Carissa15
Matrix a)\[\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]\]
jim_thompson5910
  • jim_thompson5910
how far did you get with the eigenvalues of \[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\]

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Loser66
  • Loser66
What is the question?
Carissa15
  • Carissa15
I really haven't done anything with eigenvectors or eigenvalues. Very new to me. Seems though if I
jim_thompson5910
  • jim_thompson5910
are you familiar with the idea that you need to find the determinant of the matrix \(\LARGE A - \lambda I\) to get the eigenvalues?
Carissa15
  • Carissa15
\[\det( \lambda \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]-\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right])=0\]
jim_thompson5910
  • jim_thompson5910
it should be the other way around A - (lambda)*I not (lambda)*I - A
jim_thompson5910
  • jim_thompson5910
so 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
  • Loser66
@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
  • jim_thompson5910
oh I did not know that
jim_thompson5910
  • jim_thompson5910
well however you solve the equation, what values of lambda do you get?
Carissa15
  • Carissa15
just working through now
Carissa15
  • Carissa15
I get \[\lambda^2-4\lambda-45=0\] not sure where to go from there
jim_thompson5910
  • jim_thompson5910
use the quadratic formula to solve for lambda
jim_thompson5910
  • jim_thompson5910
you can also factor the left side if you want
Carissa15
  • Carissa15
is the quadratic formula \[\lambda^2-\lambda-4=0\]?
jim_thompson5910
  • jim_thompson5910
if you factored, you'll find that \[\Large \lambda^2-4\lambda-45=0\] turns into \[\Large (\lambda-9)(\lambda+5)=0\]
jim_thompson5910
  • jim_thompson5910
so you agree that \[\Large \lambda = 9 \text{ or } \lambda = -5\] or no?
Carissa15
  • Carissa15
yes
Carissa15
  • Carissa15
so these are the eigenvalues?
jim_thompson5910
  • jim_thompson5910
correct
Carissa15
  • Carissa15
I am not sure how to now find eigenectors using these
jim_thompson5910
  • jim_thompson5910
\[\Large A\vec{v} = \lambda \vec{v}\] \[\Large A\vec{v} - \lambda \vec{v}=0\] \[\Large A\vec{v} - \lambda I\vec{v}=0\] \[\Large (A - \lambda I)\vec{v}=0\] 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 \[\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]\]
jim_thompson5910
  • jim_thompson5910
If lambda = 9, then \[\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]\] \[\Large \left[\begin{array}{cc|c}2-9 & 7 & 0 \\7 & 2-9 & 0\end{array}\right]\] \[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7 & -7 & 0\end{array}\right]\] row reduce the augmented matrix until it's in rref form
Carissa15
  • Carissa15
So we would end up with R1+R2 and all values zero?
jim_thompson5910
  • jim_thompson5910
you do compute R1+R2, yes but we only need to do that for one row R1+R2 = [0 0] replace R2 with that result \[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7&-7 & 0\end{array}\right]\] \[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\0&0 & 0\end{array}\right]\begin{array} \ \\ R1+R2 \rightarrow R2\end{array}\]
Carissa15
  • Carissa15
Oh yes, of course. not both oops.
jim_thompson5910
  • jim_thompson5910
the 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
  • jim_thompson5910
a 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
  • Carissa15
Would 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
  • jim_thompson5910
yes 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
  • jim_thompson5910
and yes you follow the same steps for lambda = -5
Carissa15
  • Carissa15
ok great. Thank you. I will work my way through them and see how I go. :)
jim_thompson5910
  • jim_thompson5910
ok tell me what you get
Carissa15
  • Carissa15
for lambda -5 i got\[\left[\begin{matrix}7 & 7 \\ 7 & 7\end{matrix}\right]\\] with Identity matrix
Carissa15
  • Carissa15
then did R1-R2 gives new R2 of all zeros.
Carissa15
  • Carissa15
I then got 7x-7y=0, which then ends up with the same result as lambda 9
jim_thompson5910
  • jim_thompson5910
I'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
  • jim_thompson5910
7x-7y=0 is incorrect
Carissa15
  • Carissa15
it is always x+y=0 ?
jim_thompson5910
  • jim_thompson5910
I don't know what you mean by "always", but 7x + 7y = 0 does reduce to x+y = 0
jim_thompson5910
  • jim_thompson5910
then x+y = 0 turns into y = -x when you solve for y
jim_thompson5910
  • jim_thompson5910
so \[\Large \left[\begin{array}{cc|c}x \\ y\end{array}\right]=\left[\begin{array}{cc|c}x \\ -x\end{array}\right]\]
jim_thompson5910
  • jim_thompson5910
if we let x = 1, then \(\Large \left[\begin{array}{cc|c}1 \\ -1\end{array}\right]\), and all of its scalar multiples, represent the other set of eigenvectors.
Carissa15
  • Carissa15
oh ok.
jim_thompson5910
  • jim_thompson5910
I 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
  • jim_thompson5910
same for lambda = -5
Carissa15
  • Carissa15
I have taken the same steps for matrix b) but get stuck as I cannot factor to find eigenvalues. Please help.
Carissa15
  • Carissa15
\[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\] I end up with \[\lambda^2-2\lambda-1=0\] but get stuck trying to factor to find eigenvalues
Carissa15
  • Carissa15
eigenvalues : \(\lambda = 1\pm \sqrt{2}\) 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
  • Loser66
|dw:1443535039866:dw|
Loser66
  • Loser66
then 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 \(1+\sqrt2\)
Loser66
  • Loser66
gtg. good luck
Carissa15
  • Carissa15
Thank you everyone. Solved :)

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