Question

eigen vectors

Answer 1

i've got a system of linear DE's: \(\dot x = - x,\; \dot y = -y\) so
\[ \mathbf{ \dot x} = \left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right] \mathbf{x}\]
[where \(\mathbf{x} = [x(t), y(t)]^T\)

this is triangular matrix so read off the eigen values but here \(\lambda_{1,2} = -1\), repeated e-values, not good

but we look for the first e-vector \(\mathbf{v}\) at least: \(\left[\begin{matrix}-1-(-1) & 0 \\ 0 & -1-(-1)\end{matrix}\right] \left[\begin{matrix}v_1 \\ v_2\end{matrix}\right] = \mathbf{O}\)

even worse.

the computer agrees with \(\lambda\)'s and says e-vectors are: \([v_1, v_2] = [0,1]^T, \; [1,0]^T\)

in the name of sanity, how do i get these e-vectors from this system ???!!!!

Answer 2

By definition, \(x\) is an eigenvector for the corresponding eigenvalue \(\lambda\), if it satisfies the euation \(Ax = \lambda x\)

Answer 3

plugin \(\lambda=-1\) and rearranging gives :

\[\begin{bmatrix}0&0\\0&0\end{bmatrix}\mathbf{x}=\mathbf{0}\]

Answer 4

Looks like we can solve this as independent separable equations,
\[\frac{dx}{dt}=-x\]\[x=C_1e^{-t}\]

Answer 5

yeah the system is decoupled already

Answer 6

\[\begin{bmatrix}0&0\\0&0\end{bmatrix}\mathbf{x}=\mathbf{0}\]

any nonzero vector that satisfies above equation is an eigenvector.
since every vector satisfies above equation, you can pick any two indepdent vectors

Answer 7

that is exactly what i needed ... and spent a few hours trying to find

i am now going outside to kick the wheelie bin.

Answer 8

Hahaha hey look on the bright side, I bet you're not gonna forget this now! xD

Answer 9

@Empty exactly, mate! i tell myself this all the time. if you breeze through life, you'll learn nothing. if you constantly make an retriceof yourself, you will be as wise as Solomon in the end ;-))

Answer 10

oh "retriceof" = _a_r_s_e

Answer 11

While you're at these cool stuff, you may want to review LA
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-21-eigenvalues-and-eigenvectors/

i review these once in a while.... gilbert strang is amazing !

Answer 12

and i now understand this too, so i'll put it here for posterity

**2 linearly independent e-vectors** only happens in 2-D land if:

\[\mathbf{A} = \alpha \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] = \alpha \mathbf{I} \quad \alpha \ne 0\]

[or am i about to get even "wiser !!??!!]

Answer 13

brilliant link, ganesh. i will watch as i have mucho to learn.

heard of him before, mate of mine swears by this guy and he does seriously funky CFD ... professionally.

Answer 14

That doesn't look right... let me think of a simple counter example...

Answer 15

cool!

Answer 16

If the eigenvalues are distinct, then the eigenvectors are also independent.

So we may take any matrix that has distinct eigenvalues

Answer 17

you must be knowing something about eigenvalues of a triangular matrix ?

Answer 18

ah

i was thinking "repeated eigenvalues". purely that scenario.

when you can have 1 independent, or 2 independent

let me put a link up as it will at least clarify what i am actually trying to do.....

Answer 19

I think it's more like a 2 dimensional eigenspace. So there's a whole plane of eigenvectors that have the same eigenvalue instead of the usual one eigenvector. That's no problem because we can use something called 'Gram-Schmidt orthogonalization' to get two orthogonal vectors in a general case although here it's not really a big deal, any 2 linearly independent vectors will do (the first two simplest ones you can find) this isn't rocket science on this problem luckily... Or at least I hope I conveyed that somehow haha.

Answer 20

eigenvectors of \(\begin{bmatrix}a&b\\0&d\end{bmatrix}\) are independent whenever \(a\ne d\)

Answer 21

Oh ok... I got you now :)

Answer 22

http://www.cds.caltech.edu/archive/help/uploads/wiki/files/179/lecture2B.pdf

this is it - phase planes, stability etc - except i am really going to be looking at non linear DE's, though you look at at them by linearising the thing.

so you really do need the e-vectors, to classify the equilibrium points....if that makes sense

Answer 23

yes @Empty

i think i'm with you too.

and i think, when i first tried it via computer, Wolfram just pulled out the 2 easiest/best ones leaving me lost.

Answer 24

there's a massive cross over here in this analysis and the usual hoo-hah of solving 2nd order DE's with const coefficient, ie you can "cheat" and uncouple things and go don that other route

that's actually quite disorienting .....as you recognise so much but are not really that sure how they are connected.

Answer 25

gtg

i am defs going to have more of these problems ..... so i will be posting more and grateful, as usual, for you guys' helps 🍀

Answer 26

I haven't really touched this stuff too much solving systems of differential equations, like we did it a few times while taking the class but that's about it. Occasionally I'll come into one and I'd like to play around with them more.

I guess when they decouple it, they're really diagonalizing the matrix? Also, when you look at nonlinear ones I know you kinda can throw away the nonlinear part but I sorta never really understood that too much and there's this Lyapunov stuff maybe? I don't know, but I should learn I've been wanting to apply some of this sorta stuff lately.

Answer 27

yeah, empty, that's pretty much the roadmap for the near term :p

enjoying strang....