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Characteristic equation

  1. Jul 12, 2006 #1
    Let's say I'm given a DEQ: (1) [tex]y^{(n)}+a_{n-1}\cdot y^{(n-1)}+\ldots + a_{0}\cdot y=0[/tex], where y is a real function of the real variable t, for example. I could now rewrite this as a system of DEQ in matrix form (let's not discuss why I would do that): (2) [tex]x'=Ax,\quad x=(y,\ldots,y^{(n-1)})[/tex]. If I substitute in (1) y=exp(r*t), I get the characteristic equation: (3) [tex]r^n+a_{n-1} r^{n-1}+\ldots+a_{0}=0[/tex]. I could however also take the characteristic equation of my matrix A: (4) [tex]det(A-rId)=0[/tex]. It is completely clear to me why those to things lead to the very same result. What I don't see is why I end up with the exactly identical algebraic equation for the determination of r. (I do understand why I get the same results for r.)
    In the case n=2 I could of course simply do the math and see that it's true there (which is what I did). But how would I proceed for an arbitrary n?
    Thanks alot for your hints.
    Best regards...Cliowa
    Last edited: Jul 13, 2006
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  3. Jul 12, 2006 #2


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    your question is hard for me to understand. if you are asking why two equations which ahve the same roots are the same equatioon, that is elementary alkgebra.

    if you want to make a link with some matrix theorym, think about companion matrices of polynomials, rational canonical forms, and cyclical subspaces. see if that helps.

    also eigenvalues and eigenfunctions.
  4. Jul 13, 2006 #3
    Not quite. There's a tiny little more to it: Those two equations must have the same polynomial degree, which btw is the case here. You know I see this argument, but it seems somehow a little "in from behind", somewhat indirect. Let me explain: I know that when rewriting this equation (1) as a system of DEQ only the last line contains information, the rest is useless. So when I then plug in some trial function like x=exp(r*t)*(a1, a2,...,an), it follows that for a solution of (2) condition (4) must hold. As this matrix A-rId doesn't contain any other information and I get the same polynomial degree ... those must be the same two equations. My question now was: Is there a better way to see this? It seems kinda indirect to me, especially because my det()-function doesn't really do the same, mathematically.
    Of course one could argue that all rows but the last one are linearly independent, so the last one should be linearly dependent from the rest, but that could be achieved without the last row having to be zero.
    I hope I made myself clear this time. Best regards...Cliowa
  5. Jul 13, 2006 #4


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    I assume you mean, for example, writing y"+ 3y'+ y= 0 as:
    u= y' so y"= u' and the equation become u'+ 3u+ y= 0. The system of equations is y'= u and u'= -3u- y. The first equation is far from "useless". Without it you wouldn't know what u meant in the second equation!
  6. Jul 13, 2006 #5


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    do you know what a companion matrix is?
  7. Jul 13, 2006 #6
    I obviously didn't make this "rewriting as a system" clear, I'm sorry. Here's what I meant: You take as a n-column vector x the vector of your function y plus all its derivatives (in ascending order) until the (n-1)st one: [tex]x=(y,y',y'',\ldots,y^{(n-1)})^{T}.[/tex] (the order of this was wrong in my first post, now edited.). Now you differentiate this: [tex]x'=(y',y'',y''',\ldots,y^{(n)})^{T} [/tex]. If you now rewrite this using matrices, then you get n-1 useless rows in your matrix, i.e. n-1 rows simply stating that y'=y', y''=y'',... The Matrix will look like this: Zeros everywhere except
    - in the first sidediagonal (right) all 1 elements
    - in the last row: the coefficients from the equation (1).

    In fact, it looks like the companion matrix for the DEQ interpreted as a polynome in y (or it's transpose if you take the definition from http://en.wikipedia.org/wiki/Companion_matrix" [Broken]).
    Last edited by a moderator: May 2, 2017
  8. Jul 13, 2006 #7
    Well, I didn't now the thing had that name! But now I looked it up, and it's what I meant (plus transposition, if necessary). So we're transferring this problem to linear algebra (which is what I was hoping for). So there's left one thing to clear: Why is the characteristic polynome of this companion matrix the same as the polynome (in this case the DEQ) itself? What's a clever way to see this (for example some clever development)?
  9. Jul 13, 2006 #8


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    look uo how you get the companion matrix, i.e. you choose a cyclic basis. then notice that the natural basis for the solutions of this differential equation are a cyclic basis. i.e. applying d to x^n gives (a multiple of) x^(n-1),.....
  10. Jul 14, 2006 #9
    Well, I'm not really sure I understand how this works and why this is defined that way. Let me tell you what I think I understand, so you can correct me if I'm wrong. I'm seeing the companion matrix to a polynome p of degree n in t as a transition matrix from one base to another, the first base being {t, t^2, ... ,t^n}, the second being {1, t, ... , t^(n-1)}. If I then write down my polynome I get those subdiagonal 1-entries. In order to get the last column however I need to set my polynome p to zero and solve for t^n. Why is this done that way? Is this being used to find roots of some polynome (I guess so...)? As I mentioned, everything except the last row is completely independent of the coefficients of p and therefore isn't quite loaded with information. Seems strange to me.
    Thanks again...Cliowa
  11. Jul 14, 2006 #10


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    stuff and nonsense

    im not really awake so i may be wrong or imprecise here, but given a linear transofrmation T, on a space V, i.e. from V to V, finite imensional, there is a minimal polynomial p such that p(T) is identically zero on V.

    the short answer to your question is that both procedures you describe yield the minimal polynomial for the lienar transformation D (i.e. differentiation), on the space V of solutions of that differential equation.

    I.e. the original de may be viewed as giving the minimal polynomial P(D) and asking for the largest space of functions on which it is the minimal polynomial.

    On the other hand, suppose P has degree d and V has dimension d and P(x) is the minimal polynomial of D on V. Then there isa theorem that there is a "cyclic" basis for D on V, and in that basis the matrix for D is a companion matrix A for that polynomial.

    Hence the minimalpolynomial P, can be calculated as the characteristic polynomial, i.e. det(X-A).

    as for plugging in e^(xt) for y in the original equation and geting P(x) = 0, that is kind of an ad hoc way of just substituting x for y in the equation and powers for derivatives, i.e. of writing the equation as P(D).

    It is a little misleading since it suggests that all solutions of the equation hVE FORM e^ct, which they do only if the roots are distinct.

    A better way of doing it is to look at the equation as P(D) = 0, and factoring the equation as a product of powers of (D-c) for vaRIOUS C'S, not necessarily all different. i.e. a product of powers (D-c)^s for different c's.

    Then solving the equation is the same as solving each of the equations

    (D-c)^s = 0, which shows where the additional solutions come from that are not of form e^(ct).

    This amkes it look more like a jordan form approach, and now one sees perhaps that another answer to your i=uestion is that one calculation computes the minimalpolynomial from the jordan matrix and the other computes it from the companion matrix, and they give the same answer.

    i have equated mnimal and characteristic polynomials in this discussion since in this case they agree. i.e. the dimension of the null space of (D-c)^s is always s for this operator D.

    But on some spaces, and with some other operators, the null space, of T-c could be more than one. that is when the characfteristic polynomial is bigger than the nimal one, but this do3s not sem to happemn for the deriavtive operator D, since Dy = 0 always ahs one dimensional solution (on an interval).:blushing:
  12. Jul 14, 2006 #11
    But are we really dealing with a linear transformation T:V to V? Because the two basis I suggested were not basis for the same vector space. That's one point why I don't seem to understand this "cyclic" basis stuff along with why this companion matrix is built up exactly that way, and not some other.

    The trouble is I see what's happening, but I don't understand how it works out. So yes, that's the short answer, but I still don't get it.

    This is essentially using other words, right?

    Could you point me to that theorem? What is the definition of a cyclic basis? Sadly I found only very little information on these subjects in the web. Do you know a book or some web page where I could read up on all this?

    I feel I do understand how the DEQ is solved, how those "extra" solutions come in and so on. It's really the linear algebra part about those polynomials I don't get.

    So, thanks alot again. Best regards...Cliowa
  13. Jul 14, 2006 #12


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    "But are we really dealing with a linear transformation T:V to V? "

    yes. the vector space is the solution space of that equation, and T = D is differentiation, and the derivative of a solution is again a solution for that equation.
    Last edited: Jul 14, 2006
  14. Jul 14, 2006 #13


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    a cyclic basis for a vector space V, with respect to a linear map T on V, i s a basis of the form, {w, Tw,T^2w,...,T^(n-1)w,}

    where T^n w = Q(T)w, Q is some polynomial of degree n-1 or less in T, i.e. T^n w depends linearly on the elements of the basis above.

    in such a basis the matrix of T is a companion matrix.

    e.g. if T^n = 0, the companion matrix has all zeroes in the right hand column, and the other columns have a single one below the diagonal.

    if your polynomial is (X-c)^n, then the de (D-c)^n = 0, has a solution space with basis e^cx, xe^cx, x^2e^cx,....

    and this is almost a cyclic basis, not for D but for D-c, if written backwrads, i.e. (D-c) xe^cx = e^cx +ce^cx - ce^cx = e^cx.

    in the same way, (D-c) [x^r e^cx] = rx^(r-1) e^cx, so we have to divide by r to get a cyclic basis.

    anyway, i don't have time to teach a whole cousre in linear constant coeff de's here, but this is all linbear algebra. and i think the linear algebra concepts of companion abses ans \jordan bases and so on were probably invented to make sense of this diff equation.

    you might not get the answer to the exact question you are qskibng but you might want to download my notes on linear algebra, the primer, from my webpage.

    oh yes, the thm that every finite abelian group is a sum of cyclic grouops has an analog that every finite dimensional vector space acted on by a linear map T, decomposes into subspaces that each have a cyclic basis.

    that mens essentially that the vector space is isomrphic to a quotient space of form k[X]/)P) where P is apolynomial. hence a vector basis is given by the elments 1,x,x^2,.... up to n-1 where n is the orderof the polynomial p.

    if you think of multiplication by x as the same as appklying the operator T, this is a cyclic basis.

    this thm is proved in my notes.
  15. Jul 14, 2006 #14


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    the proof is just diagonalizing a matrix, but in my notes there the proof is different, by induction. the matrix proof is in my math 845 notes also on my webpage.
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