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I Matrix Equation -- clarification about solving a system

  1. Aug 1, 2017 #1

    joshmccraney

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    Hi PF!

    Just want to make sure I'm not crazy: if we're solving a system ##K a = \sigma^2 M a## where ##K## and ##M## are ##n\times n## matrices, ##a## an ##n\times 1## vector and ##\sigma## a scalar, then ##a## is unnecessary, and all we really need to solve is ##K=\sigma^2 M##, right?
     
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  3. Aug 1, 2017 #2

    fresh_42

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    Depends on the intention, i.e. what is meant. ##K=\sigma^2 M## describes the entire kernel ##\mathcal{K}=\operatorname{ker} (K-\sigma^2M)## and ##Ka=\sigma^2 Ma## means a certain vector ##a \in \mathcal{K}##.
     
  4. Aug 1, 2017 #3

    joshmccraney

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    What do you mean by "describes entire kernel"? I don't know ##a## and am trying to determine it from the equation ##Ka=\sigma^2Ma## but it seems impossible to determine a non-trivial ##a## from this. Your thoughts?
     
  5. Aug 1, 2017 #4

    fresh_42

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    Let's say ##\varphi = K - \sigma^2M## for short and my laziness.

    I don't know where it came from, so one possibility could have been: A statement where a vector ##a## occurs at some point and a condition ##\varphi(a)=0## is derived later. This way the ##a## already was there before the condition showed up.

    If you want to solve ##\varphi(a)=0## for all possible ##a## then this means to calculate a basis of the vector space ##\operatorname{ker}\varphi = \{ a \,\vert \, \varphi(a)=0 \}## and a certain ##a## can be expressed in this basis.

    One possibility is, that ##\operatorname{ker}\varphi = \{0\}## in which case ##a=0## is a consequence, which means ##\varphi## is injective. So whether there are non-trivial solutions ##a## is equivalent to whether ##\operatorname{dim} \operatorname{ker} \varphi## is zero or positive.

    You've asked, whether there is a difference. And the answer is: given ##a## with ##a \in \operatorname{ker}\varphi## (formula with ##a##) is formally different from ##\operatorname{ker}\varphi## (formula without referencing ##a##) which describes all possible ##a##.
     
  6. Aug 5, 2017 #5

    FactChecker

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    Is ##K a = \sigma^2 M a## true for all vectors a or just for one vector a?
     
  7. Aug 5, 2017 #6

    joshmccraney

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    It's an eigenvalue problem, only instead of ##M=I## it's something different. To solve I'm taking ##\det(K-\sigma^2 M)=0##. Sound right?
     
  8. Aug 5, 2017 #7

    FactChecker

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    That sounds wrong to me, but maybe I'm missing something. In the first post, it sounds like you are asking if K and σ2M agree for one vector, a, then determine if K and σ2M are identical. But K and σ2M either are identical or they are not, there is no solving to do.
     
  9. Aug 5, 2017 #8

    fresh_42

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    Well, at least it can be used to decide whether there is a non-trivial ##a## or not. But you won't find it this way.
     
  10. Aug 5, 2017 #9

    joshmccraney

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    Sorry, let me be clear: I'm trying to solve the eigenvalue problem ##K_{ij}a_j=\sigma^2M_{ij}a_j##. I know what ##M## and ##K## are, so I need to determine the eigenvalue ##\sigma^2## and the eigenvector ##a##. I thought solving this algebraic system is equivalent to $$K_{ij}a_j=\sigma^2M_{ij}a_j \implies\\
    Ka=\sigma^2 M a\implies\\
    (K-\sigma^2M)a=0.$$
    Then we solve for ##\sigma^2## by solving ##\det(K-\sigma^2M)=0##. Isn't this standard and correct?
     
  11. Aug 5, 2017 #10

    fresh_42

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    ##\operatorname{det}(K-\sigma^2M)\neq 0## means, only ##a=0## fulfills the equation ##Ka=\sigma^2 Ma\,.##
    ##\operatorname{det}(K-\sigma^2M) = 0## means, there is an ##a_\sigma \neq 0## with ##Ka_\sigma=\sigma^2 Ma_\sigma\,.##
    It doesn't tell how many linear independent ##a_\sigma## there are or which.
     
  12. Aug 6, 2017 #11

    joshmccraney

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    Right, but that is how you find the eigenvalues, right? To find eigenvectors corresponding to the eigenvalues I'd then solve ##(K-\sigma^2 M)a=0##, right?
     
  13. Aug 6, 2017 #12

    fresh_42

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    I'm not sure whether I would use this method to determine possible values for ##\sigma^2##, as it looks as one could get an unpleasant polynomial in ##\sigma^2## which might not be easy to find roots for. Maybe it would be easier to use ##\sigma^2## to parameterize the solutions of the linear equations ##(K-\sigma^2 M)a=0## and do everything in one step. But basically, yes.

    Another point is, that I have some difficulties to call them eigenvalues and eigenvectors. Of what? As I understand it, you are only interested in eigenvectors ##a_\sigma## to the eigenvalue ##0## of ##\varphi_\sigma = K-\sigma^2 M##, resp. possible pairs ##(\sigma,a_\sigma)## or ##(\sigma,\operatorname{ker}\varphi_\sigma)\,.##
     
  14. Aug 6, 2017 #13

    joshmccraney

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    Yea, the polynomial could be ugly but I'm letting Mathematica crank through it. The eigenvalues/vectors correspond to perturbations of a capillary surface. The equation ##Ka=\sigma^2Ma## is an integro-differential equation, though I've not defined those matrices here for convenience.
     
  15. Aug 6, 2017 #14

    fresh_42

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    The difficulty with those equations is, that numerically derived roots aren't stable solutions, i.e. if we wobble a bit at ##\sigma##, we'll get stuck with ##a=0## since the solutions different from zero are dense, and the solutions are not.
     
  16. Aug 6, 2017 #15

    joshmccraney

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    Hmmmm so how would you do it? You mentioned parameterize ##\sigma^2##; can you elaborate?

    Also, if I were solving ##Ka=\sigma^2Ma## couldn't I instead do ##M^{-1}Ka=\sigma^2 a##, which can be recast as ##(M^{-1}K-\sigma^2I)a=0##, which amounts to finding the eigenvectors/values of the matrix ##M^{-1}K##?
     
  17. Aug 6, 2017 #16

    fresh_42

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    I thought of something like Gauß elimination to solve for ##(K - \sigma^2 M)x = 0## and keeping ##\sigma^2## in the process as variable, if this is possible. Then you see at the result for which values of ##\sigma^2## there are non-trivial solutions and for which there are none. But I'm not really good in numerical analysis, so maybe the same numbers which make the determinant ugly as you said, will also make the description of the solutions by matrix operations ugly. And maybe it doesn't matter for your problem. I only wanted to say, that ##\det (K-\sigma^2 M) = 0## is a straight, which means a little disturbance and you're no longer on it; but then ##\det (K-\sigma^2 M) \neq 0## and ##a=0## is the only solution.
     
  18. Aug 6, 2017 #17

    joshmccraney

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    Thanks for your insight! I appreciate it!
     
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