Question

find the eigenvector:
http://gyazo.com/e0e91a80e36cbe597a48b82d2e909313

Answer 1

i tried it but i dont get why, but im stuck :|

Answer 2

Well, let's try it and see what happens, shall we?
\[\Large \left[\begin{matrix}2&0&0\\1&1&0\\1&2&3\end{matrix}\right]\left[\begin{matrix}x\\y\\z\end{matrix}\right]=3\left[\begin{matrix}x\\y\\z\end{matrix}\right]\]

Answer 3

Sure, so what did you do when you attempted to find the eigenvector?

Answer 4

Whoops... cat's out of the bag, my bad....

Answer 5

http://gyazo.com/9103da595959b17e8e752db72480f24b

Answer 6

im unable to get the zeroes at bottom

Answer 7

What you're doing with that process is finding the eigenvalues (if I'm not mistaken)
but such processes are unnecessary when you already have an eigenvalue and you're simply asked to find eigenvectors which correspond to it~

Answer 8

im trying to find the eigenvector with the eigenvalue given..

Answer 9

am i allowed to switch the last column (0,0,0) to be the last row?

Answer 10

Never lose sight of the actual definition of an eigenvalue/eigenvector.
Let \(\lambda\) be some scalar and \(\vec v\) a vector, and A be a matrix.

If \(\large A\vec v=\lambda \vec v\), then \(\lambda\) is an eigenvalue of A with eigenvector \(\vec v\)

Answer 11

Sure, so the eigenvector is just simply the sum of the columns in your matrix. Here's the algebra:

\[Ax=\lambda x\]\[Ax-\lambda x=0\]\[Ax-\lambda I x=0\]\[(A-\lambda I) x=0\]\[\det(A-\lambda I)=0\]

Now since that was already done for you, you can see that the final thing shows that you will get a noninvertible matrix, right? If it was invertible, the determinant wouldn't be equal to zero!

Check that adding up the columns gives you an eigenvector by just simply multiplying it and seeing if you can factor out a constant.

Answer 12

yeah.but im an unable to get 0's at bottom row..

Answer 13

Yeah, but why do you want 0's at the bottom row? That doesn't really get you anywhere.

Answer 14

so i can do a substitution \(z=a\) ?

Answer 15

in my lecture notes, the bottom row is all zeroes..

Answer 16

http://www.wolframalpha.com/input/?i=row+reduce++%28+%282-3%2C0%2C0%29%2C+%281%2C1-3%2C0%29+%2C%281%2C2%2C3-3%29%29

Answer 17

Wha...? No, stop. They might be zeroes some times. There's no reason for them to always be.

Answer 18

seems you're stuck at elimination step ? @mathsnerd101 ?

Answer 19

yee thanks for the link!

Answer 20

There's really no reason to be doing this. We already know the determinant of this matrix is zero because it's an eigenvalue. This is simply showing that indeed it is a linearly dependent set, but it doesn't do anything towards finding the eigenvector itself.

Answer 21

well, how can i find the eigenvector without getting the 0's at the bottom?

Answer 22

Sure, so the eigenvector is just simply the sum of the columns in your matrix.

Answer 23

Might I suggest just doing this?
\[\Large \left[\begin{matrix}2&0&0\\1&1&0\\1&2&3\end{matrix}\right]\left[\begin{matrix}x\\y\\z\end{matrix}\right]=3\left[\begin{matrix}x\\y\\z\end{matrix}\right]\]

Answer 24

then what else?

Answer 25

<shrugs>

Evaluate the left-hand side, using matrix multiplication?

Answer 26

\[\Large \left[\begin{matrix}2x\\x+y\\x+2y+3z\end{matrix}\right]=\left[\begin{matrix}3x\\3y\\3z\end{matrix}\right]\]

Now it's just a system of equations ^_^

Answer 27

solve for x y and z?

Answer 28

Do try :D

Answer 29

I suppose once you understand that, you can see how simply adding the columns of that matrix gives you an eigenvector. There are infinitely many eigenvectors so it doesn't really matter which one you get. You can always divide/multiply it by any scalar kind of like simplifying a fraction until you get what you want.

Answer 30

I actually didn't know anything about how adding columns gives an eigenvector, my understanding of matrices is quite minimal, hence, me sticking to definitions ^^

Answer 31

Ahh, if you're curious you can integrate stuff like e^x*sinx instead of by parts twice and cutting itself off or whatever, instead you can solve it by simply inverting a 2x2 matrix. Linear algebra is awesome!

Answer 32

I still would rather steer clear of it if I can avoid it :D

Answer 33

@mathsnerd101 do you have your answer?