Question

Eigenvectors / eigenrooms (Question in comment)

Answer 1

So I need a bit of help understanding how to find the eigenroom. So I understand that you take the "independent" variables and describe them as a vector, of the dependent ones. So like \[A=\left[\begin{matrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{matrix}\right]\] Would have \[u=\left[\begin{matrix} 1\\ 1 \\ 0\end{matrix}\right]\]
Because I set up the equations \[x_1-x_2=0 -> x_1=x_2\]\[x_2=x_2\]
But what if I have \[A=\left[\begin{matrix} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\] How would i find the two vectors? that would give me the equations: \[x_1+x_2-x_3=0\]\[x_2=x_2\]\[x_3=x_3\] Right? And how do I find the two vectors from this?

Answer 2

what is the actual matrix ?

Answer 3

Don't you mean the matrix \(\left[\begin{matrix} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\) represents \(A-\lambda I\) ?

Answer 4

First I have, \[A=\left[\begin{matrix} 5 & -1 & 1 \\ -1 & 5 & 1 \\ 0 & 0 & 6\end{matrix}\right]\]
Then I have found the 3 different eigenvalues, by taking the determinant of
\[A=\left[\begin{matrix} 5-l & -1 & 1 \\ -1 & 5-l & 1 \\ 0 & 0 & 6-l\end{matrix}\right]\] And I got \[l=4,6,6\] Now when I insert \[l=6\] I get
\[A=\left[\begin{matrix} -1 & -1 & 1 \\ -1 & -1 & 1 \\ 0 & 0 & 0\end{matrix}\right]\] And then from gaus elimination I get: \[A=\left[\begin{matrix} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\]

Answer 5

l is supposed to be "Lambda"

Answer 6

Now it makes sense... gimme a minute to go through your work :)

Answer 7

Yea, sure, thanks :)

Answer 8

you may use `\lambda` to write \(\lambda \)

Answer 9

Oh yea okay thx ;)

Answer 10

for \(\lambda = 6\), we do get two independent eigenvectors

Answer 11

Yes, I am having trouble finding out how to find these, because I have \[x_1+x_2-x_3=0\]\[x_2=x_2\]\[x_3=x_3\], and I dont know how to describe those into 2 different vectors.

Answer 12

Easy, simply let \((x_2, x_3) = (1, 0)\) to get one eigen vector

Answer 13

Again let \((x_2, x_3) = (0, 1)\) to get another eigen vector

Answer 14

Whats an Eigenroom?? the set of eigen vectors that define different paralleipeds?

Answer 15

\[x_1+x_2-x_3=0\]

plugin \((x_2,x_3)=(1,0)\) in above equation and solve \(x_1\)

Answer 16

These special values guarantee that you get "independent" eigenvectors

Answer 17

\[x_1+x_2-x_3=0\] If I insert \[(x_2,x_3)=(1,0)\] do I then get \[x_1=-x_2\]?

Answer 18

plugin \(x_2 = 1\) and \(x_3=0\)

Answer 19

\[x_1+1=0\]\[x_1=-1\]

Answer 20

\(x_1+1-0 = 0\)

\(x_1=?\)

Answer 21

Yep, so one eigen vector is \((-1,1,0)\)

Answer 22

Let \((x_2, x_3) = (0, 1)\) to get another independent eigen vector

Answer 23

Hmm. According to the test results (It is exam practice, where we have the answers) The answers are \[Span[(1,0,1),(0,1,1)]\]

Answer 24

http://www.wolframalpha.com/input/?i=eigenvectors+%7B%7B5%2C-1%2C1%7D%2C%7B-1%2C5%2C1%7D%2C%7B0%2C0%2C6%7D%7D

Answer 25

I have 4 answers, the 4th says: Neither one of the above.

Answer 26

And it says the answer is [3]

Answer 27

Yes, answer is indeed [3]

Answer 28

Our answer is also going to match with [3]

Answer 29

Can you finish it off ?

Answer 30

find the other eigenvector, don't wry about the options yet

Answer 31

Yea, but wolframalpha / what we did gives 3 vectors?

Answer 32

Ok

Answer 33

So next I would insert \[(x_2,x_3)=(0,1)\] \[x_1+0-1=0\]\[x_1=1\]

Answer 34

wolfram's 3 eigenvectors correspond to ALL the eigenvalues

your test answer is referring to just the eigenvectors corresponding to the eigenvalue 6

Answer 35

Yes, it was just before we got \[(-1,1,0)\] Which I assume comes from when \[\lambda =4\]

Answer 36

Yes. So, what are the eigenvectors corresponding to the eigenvalue 6 ?

Answer 37

So we get:\[(1,0,1)\] and \[(-1,1,0)\]?

Answer 38

Yes. that looks good.

Answer 39

Now lets look at options

Answer 40

Looking at options, using these vectors, I would choose [4], none of the above.

Answer 41

Not so fast. Recall that if \(v_1\) and \(v_2\) are any two eigenvectors, then any linear combination \(c_1v_1 + c_2v_2\) is also an eigenvector

Answer 42

\[(1,0,1)\] and \[(-1,1,0)\]

Add these and look at options again

Answer 43

That gives \[(0,1,1)\]

Answer 44

Yes, \((0, 1, 1)\) is also an eigenvector

Answer 45

So yea, thats how they got it.

Answer 46

There is nothing special about the eigenvectors that we have got earlier.

Any two "independent" eigenvectors will do the job

Answer 47

So would this also work in another case, inserting \[(x_1,x_2)=(0,1)\] and \[(x_1,x_2)=(1,0)\]?

Answer 48

Which case ?

Answer 49

Well I dont know, just like in general

Answer 50

If am to find to eigenvectors

Answer 51

two*

Answer 52

You plugin special values for "free variables"

Answer 53

In our case, \(x_2\) and \(x_3\) are free variables because they don't have pivots

Answer 54

Yeap, and these "special values" are they always (0,1) and (1,0)

Answer 55

Or would they correspond to the values found in the other lines, like in our we have nothing. So we only have the equations \[x_2=x_2\] and \[x_3=x_3\]

Answer 56

You could use anything you like. But there is no guarantee that you get "indepdent" eigenvectors when you use random values/

Answer 57

ohhh, so you insert a "smart values" that eliminates one of the free variables?

Answer 58

If we use the special values (0, 1) and (1, 0) etc for free variables, we always get "independent" eigenvectors

Answer 59

yea, okay think I got it :)

Answer 60

Thank you so much :)

Answer 61

np :) one sec, wait

Answer 62

sure

Answer 63

try going thru this video when free
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-7-solving-ax-0-pivot-variables-special-solutions/

it expliains in detail how to cookup special solutions

Answer 64

Thanks :)

Answer 65

finding eigenvectors require you to solve the system :

\[\left[\begin{matrix} 5-l & -1 & 1 \\ -1 & 5-l & 1 \\ 0 & 0 & 6-l\end{matrix}\right]\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=0\]

that video explains everythign about nullspace
Enjoy...