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hamza_b23

how do we find eigenvectors of any matrix where detA =0?

  • one year ago
  • one year ago

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  1. hamza_b23
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    i mean det (A -(lamda)I)= 0 say for example A = (1 2;-1 4)

    • one year ago
  2. hamza_b23
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    i know the eigen values are lamda = 3 and lamda =2 but how to find eigenvectors

    • one year ago
  3. hamza_b23
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    @KingGeorge

    • one year ago
  4. KingGeorge
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    If you have a matrix with a determinant of 0, I'm not sure what to do. However, for your example, let \[A=\begin{bmatrix} 1&2\\-1&4\end{bmatrix}\]Then we look at the determinant of \(A-\lambda I_2\). So we have \[\det\left(\begin{bmatrix} 1&2\\-1&4\end{bmatrix}-\begin{bmatrix} \lambda&0\\0&\lambda\end{bmatrix}\right) \\ =\det\left(\begin{bmatrix} 1-\lambda&2\\-1&4-\lambda\end{bmatrix}\right) \\ =(1-\lambda)(4-\lambda)-(2)(-1)\]And set this equal to 0.

    • one year ago
  5. hamza_b23
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    yeah i know how to find the eigenvalues completely like i am good at that but it asks what are the eigen vectors thats where i get confused

    • one year ago
  6. KingGeorge
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    To be honest, I'm not too great at eigenvectors either.

    • one year ago
  7. hamza_b23
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    hmm true they are a pain. like i am reading the theorem you have to sub it back into (A - (lamda)I)x =0 and then solve for x using those lamdas but i reduce my matrix down for both the lamdas and i get x1 = x2 but wth does that mean

    • one year ago
  8. DanielxAK
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    You've pretty much solved one of the eigenvectors already. You just simplified it too much. x1 = x2 means x1 - x2 = 0. What values of x1 and x2 will solve that equation? Eigenvectors are not unique, so there's infinitely many that will work in this case. Then, your eigenvector is [x1 x2]. Note: Eigenvalues are unique, so your eigenvector only corresponds to that eigenvalue, not both. You'll have to plug in the next eigenvalue to get your other eigenvector.

    • one year ago
  9. KingGeorge
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    So if we have \(\lambda=2,3\), then we have \(A=2I_2 x\) or \(A=3I_2 x\). Let's start with 3. That gives us\[\begin{bmatrix} 1&2\\-1&4\end{bmatrix}\begin{bmatrix} x_1\\x_2\end{bmatrix}=3\begin{bmatrix} x_1\\x_2\end{bmatrix}\]Solving for this, we get that \(x_1=x_2\). So I guess that means that your eigenvector for \(\lambda=3\) is just any vector where both \(x_1\) and \(x_2\) are the same.

    • one year ago
  10. KingGeorge
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    If you solve for \(\lambda=2\) instead, you get that \(x_1=2x_2\). So in this case, your eigenvector for \(\lambda=2\) is any vector where \(x_2\) is twice \(x_1\).

    • one year ago
  11. hamza_b23
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    hmm ya thats what it said but i wasnt getting the right answer do they use any x values like we can use any number right?

    • one year ago
  12. KingGeorge
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    You should be able to use any numbers you want, unless I'm grossly misunderstanding eigenvectors.

    • one year ago
  13. DanielxAK
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    Right. So, in the example where lamda is equal to 3, you have x1 - x2 = 0. So, [x1 x2] could be [1, 1] or [3,3] or [sqrt(2),sqrt(2)]. The simpler the better though, in my opinion.

    • one year ago
  14. hamza_b23
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    true for some reason they have [1/sqrt2 1/sqrt2] thats where i was going crazy in how they determined that

    • one year ago
  15. KingGeorge
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    They wanted a normalized vector apparently (length of 1). Anyways, I've got to get going. Good luck.

    • one year ago
  16. hamza_b23
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    ooo alright thanks a lot though

    • one year ago
  17. hamza_b23
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    but @DanielxAK to get a normalized eigen vector we find the main eigenvector and normalize it?

    • one year ago
  18. DanielxAK
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    Yes, if x is your vector, then to normalize it, you divide it by the norm: x/||x||. The norm of x (||x||) is just the distance equation. (So, sqrt(x1^2 + x2^2) = ||x||) Also, note that the normalized eigenvector is unique. That's probably one reason why they're asking you to normalize it.

    • one year ago
  19. hamza_b23
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    oo true probably thanks a lot man, just another question do you know anything about symmetric matrices?

    • one year ago
  20. DanielxAK
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    A bit. What's your question?

    • one year ago
  21. hamza_b23
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    how do we show something is symmetric like i know A^T = A but its talking about some orthagonal stuff where A = CDC^T

    • one year ago
  22. DanielxAK
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    Can you state the whole question? Is it asking you to show A^T = A when A = CDC^T? I'm guessing C is an orthogonal matrix and D is a diagonal matrix?

    • one year ago
  23. hamza_b23
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    yeahh it isnt a question my TA just said just know something about that because it may be on the test

    • one year ago
  24. hamza_b23
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    is this like a spectral decomposition?

    • one year ago
  25. DanielxAK
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    Yes, sort of. It's a very specific type of decomposition. Matrices that have certain properties are easier to decompose. Decomposition is important in computing solutions to matrices. Some decompositions are more ideal than others. Anyway, If A = CDC*, then A* = (CDC*)* = (C*)*D*C*= CD*C* = CDC* = A. (D* = D since D is diagonal). Does that make sense?

    • one year ago
  26. hamza_b23
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    hmm not quite

    • one year ago
  27. hamza_b23
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    i didnt know (C*DC)* = (C*)*D*C* like i thought it was (AB)* = B*A* ?

    • one year ago
  28. DanielxAK
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    Correct. But you have three matrices instead of two. So, in the case of two matrices, you're reversing the order. The same is true for three.

    • one year ago
  29. hamza_b23
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    hmm true do you mind showing me an example of spectral decomposition if you dont mind ?

    • one year ago
  30. hamza_b23
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    i have A = ( 3 6 -1; 6 9 4; -1 4 3)

    • one year ago
  31. DanielxAK
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    I'm not going to work it out by hand. That'd be quite a bit of work for a 3x3 matrix. I did it on Matlab really quick though. So, you'd have it of the form: A = C*D*C' Where, roughly, C = [ -0.4551 0.5802 -0.6754 -0.8460 -0.0451 0.5313 -0.2778 -0.8132 -0.5114] D = [ 13.5417 0 0 0 3.9353 0 0 0 -2.4770]

    • one year ago
  32. hamza_b23
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    ok sounds good and C= the normalized matrix for A?

    • one year ago
  33. DanielxAK
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    C is your unitary (or orthogonal) matrix. Spectral decomposition is also known as eigenvalue decomposition. If you worked out the eigenvalues of that matrix, you would see that diagonal of D makes up your eigenvalues and each column of C makes up your normalized eigenvectors (which correspond to its eigenvalue). For example, v = -0.4551 -0.8460 -0.2778 corresponds to the eigenvalue 13.5417.

    • one year ago
  34. hamza_b23
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    ahh okayy i have to keep on practicing thenn i apparently have to do this by hand tommorow lol

    • one year ago
  35. DanielxAK
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    For a 3x3? I hope not. That's prone to error if done by hand. A 2x2 isn't too bad, but it gets much worse past that.

    • one year ago
  36. hamza_b23
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    hmm truee just one final question, i have found the eigenvalues its just the vectors that throw me off

    • one year ago
  37. hamza_b23
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    I have A = (3 1 1; 1 0 2; 1 2 0)

    • one year ago
  38. hamza_b23
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    and i found the eigenvalues to be lamda1=1, lamda2=4, lamda3 = -2

    • one year ago
  39. hamza_b23
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    and the question states that. Find the normalized eigenvectors and use them as columns in orthagonal matrix C

    • one year ago
  40. DanielxAK
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    Alright. First, build your matrix D using your eigenvalues since you know those. Then, compute the eigenvector for each eigenvalue. Normalize the eigenvectors. Then, to build your C, match the column of C with the diagonal of D. So, the eigenvalue of your first diagonal of D should match your eigenvector which is the first column of C.

    • one year ago
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