IrishBoy123
  • IrishBoy123
eigen vectors
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
IrishBoy123
  • IrishBoy123
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 ???!!!!
ganeshie8
  • ganeshie8
By definition, \(x\) is an eigenvector for the corresponding eigenvalue \(\lambda\), if it satisfies the euation \(Ax = \lambda x\)
ganeshie8
  • ganeshie8
plugin \(\lambda=-1\) and rearranging gives : \[\begin{bmatrix}0&0\\0&0\end{bmatrix}\mathbf{x}=\mathbf{0}\]

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Empty
  • Empty
Looks like we can solve this as independent separable equations, \[\frac{dx}{dt}=-x\]\[x=C_1e^{-t}\]
ganeshie8
  • ganeshie8
yeah the system is decoupled already
ganeshie8
  • ganeshie8
\[\begin{bmatrix}0&0\\0&0\end{bmatrix}\mathbf{x}=\mathbf{0}\] any nonzero vector that satisfies above equation is an eigenvector. since every vector satisfies above equation, you can pick any two indepdent vectors
IrishBoy123
  • IrishBoy123
that is exactly what i needed ... and spent a few hours trying to find i am now going outside to kick the wheelie bin.
Empty
  • Empty
Hahaha hey look on the bright side, I bet you're not gonna forget this now! xD
IrishBoy123
  • IrishBoy123
@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 ;-))
IrishBoy123
  • IrishBoy123
oh "retriceof" = _a_r_s_e
ganeshie8
  • ganeshie8
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 !
IrishBoy123
  • IrishBoy123
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 !!??!!]
IrishBoy123
  • IrishBoy123
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.
ganeshie8
  • ganeshie8
That doesn't look right... let me think of a simple counter example...
IrishBoy123
  • IrishBoy123
cool!
ganeshie8
  • ganeshie8
If the eigenvalues are distinct, then the eigenvectors are also independent. So we may take any matrix that has distinct eigenvalues
ganeshie8
  • ganeshie8
you must be knowing something about eigenvalues of a triangular matrix ?
IrishBoy123
  • IrishBoy123
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.....
Empty
  • Empty
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.
ganeshie8
  • ganeshie8
eigenvectors of \(\begin{bmatrix}a&b\\0&d\end{bmatrix}\) are independent whenever \(a\ne d\)
ganeshie8
  • ganeshie8
Oh ok... I got you now :)
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
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.
IrishBoy123
  • IrishBoy123
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.
IrishBoy123
  • IrishBoy123
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 🍀
Empty
  • Empty
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.
IrishBoy123
  • IrishBoy123
yeah, empty, that's pretty much the roadmap for the near term :p enjoying strang....

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