Question

I have a few questions about eigenvalues and eigenvectors in matrices.

Answer 1

Matrix b}
\[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\]

Answer 2

Matrix a)\[\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]\]

Answer 3

how far did you get with the eigenvalues of \[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\]

Answer 4

What is the question?

Answer 5

I really haven't done anything with eigenvectors or eigenvalues. Very new to me. Seems though if I

Answer 6

are you familiar with the idea that you need to find the determinant of the matrix \(\LARGE A - \lambda I\) to get the eigenvalues?

Answer 7

\[\det( \lambda \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]-\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right])=0\]

Answer 8

it should be the other way around

A - (lambda)*I

not

(lambda)*I - A

Answer 9

so you need to solve

\[\Large \det( \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]-\lambda\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right])=0\]

Answer 10

@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.

Answer 11

oh I did not know that

Answer 12

well however you solve the equation, what values of lambda do you get?

Answer 13

just working through now

Answer 14

I get \[\lambda^2-4\lambda-45=0\] not sure where to go from there

Answer 15

use the quadratic formula to solve for lambda

Answer 16

you can also factor the left side if you want

Answer 17

is the quadratic formula \[\lambda^2-\lambda-4=0\]?

Answer 18

if you factored, you'll find that

\[\Large \lambda^2-4\lambda-45=0\]

turns into

\[\Large (\lambda-9)(\lambda+5)=0\]

Answer 19

so you agree that

\[\Large \lambda = 9 \text{ or } \lambda = -5\] or no?

Answer 20

yes

Answer 21

so these are the eigenvalues?

Answer 22

correct

Answer 23

I am not sure how to now find eigenectors using these

Answer 24

\[\Large A\vec{v} = \lambda \vec{v}\]

\[\Large A\vec{v} - \lambda \vec{v}=0\]

\[\Large A\vec{v} - \lambda I\vec{v}=0\]

\[\Large (A - \lambda I)\vec{v}=0\]

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

\[\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]\]

Answer 25

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

Answer 26

So we would end up with R1+R2 and all values zero?

Answer 27

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

\[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7&-7 & 0\end{array}\right]\]
\[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\0&0 & 0\end{array}\right]\begin{array} \ \\ R1+R2 \rightarrow R2\end{array}\]

Answer 28

Oh yes, of course. not both oops.

Answer 29

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

Answer 30

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

Answer 31

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?

Answer 32

yes I've had teachers who would assign values to x. 1 is the smallest possible to allow for other multiples to be generated

Answer 33

and yes you follow the same steps for lambda = -5

Answer 34

ok great. Thank you. I will work my way through them and see how I go. :)

Answer 35

ok tell me what you get

Answer 36

for lambda -5 i got\[\left[\begin{matrix}7 & 7 \\ 7 & 7\end{matrix}\right]\\] with Identity matrix

Answer 37

then did R1-R2 gives new R2 of all zeros.

Answer 38

I then got 7x-7y=0, which then ends up with the same result as lambda 9

Answer 39

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

Answer 40

7x-7y=0 is incorrect

Answer 41

it is always x+y=0 ?

Answer 42

I don't know what you mean by "always", but 7x + 7y = 0 does reduce to x+y = 0

Answer 43

then x+y = 0 turns into y = -x when you solve for y

Answer 44

so

\[\Large \left[\begin{array}{cc|c}x \\ y\end{array}\right]=\left[\begin{array}{cc|c}x \\ -x\end{array}\right]\]

Answer 45

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.

Answer 46

oh ok.

Answer 47

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)

Answer 48

same for lambda = -5

Answer 49

I have taken the same steps for matrix b) but get stuck as I cannot factor to find eigenvalues. Please help.

Answer 50

\[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\] I end up with \[\lambda^2-2\lambda-1=0\] but get stuck trying to factor to find eigenvalues

Answer 51

eigenvalues :
\(\lambda = 1\pm \sqrt{2}\)

eigenvectors :
\[\begin{bmatrix} 3-(1+\sqrt{2})&-2\\1&-1-(1+\sqrt{2})\end{bmatrix} \begin{pmatrix} x_1\\x_2\end{pmatrix}= 0\]