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Linear algebra problem

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data
    If A is an n x n matrix over R such that A^3 = A + 1, prove that det(A) > 0 .

    2. Relevant equations


    3. The attempt at a solution
    So, what I've done is factor the expression to get A(A+1)(A-1) = 1, then taking the determinant of both sides, I get det(A)det(A+1)det(A-1) = 1. I thought that maybe I could argue that maybe if det(A) < 0, then this would lead to a contradiction, but that doesn't seem to be going anywhere.
    Should I try to solve it without the explicit use of determinants?
     
  2. jcsd
  3. Aug 6, 2017 #2

    StoneTemplePython

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    Gold Member

    What are the eigenvalues of ##\mathbf A## ?
     
  4. Aug 6, 2017 #3
    Wouldn't any eigenvalue have to satisfy the equation ##\lambda^3 = \lambda + 1##? Whose only real solution is approximately 1.3247?
     
  5. Aug 6, 2017 #4

    fresh_42

    Staff: Mentor

    The decomposition is a good starting point. It tells us that ##\det A = 0## is impossible, so only ##\det A < 0## or ##\det A > 0## are left.
    Now assume ##\det A < 0## and consider the real function ##f(x)=\det (A-x \cdot 1)##. Is it continuous? Do you know something about the values at certain points? What does this mean for the other points?

    Edit: I wouldn't consider eigenvalues, because you cannot be sure whether they are real or not. So eventually they don't exist.
     
  6. Aug 6, 2017 #5

    StoneTemplePython

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    Indeed. So any possible determinant is a product comprised of this real root (which is positive) raised to some natural number and the product of some conjugate pairs raised to a natural number...
     
  7. Aug 6, 2017 #6
    What can we do with ##f(x) = \det (A - xI)##? It doesn't seem to be a very malleable expression. The only two things I see is that f(0) < 0, by supposition. How does this let us conclude things about f(1) or f(-1)?
     
  8. Aug 6, 2017 #7

    fresh_42

    Staff: Mentor

    Well, you also know ##f(0)\cdot f(-1) \cdot f(1) = 1 > 0## and all are real numbers.

    Edit: You can also use @StoneTemplePython's approach with conjugate eigenvalues, if this is easier to you. My suggestion is quasi by foot using the properties of continuous real functions.
     
  9. Aug 6, 2017 #8
    Well if ##f(0) < 0## it would have to be the case that ##f(1) \cdot f(-1) < 0##, which is to say that they have different signs, which would mean that we'd have to have a zero of f(x) on the interval ##[-1,1]##, right? Is this a contradiction since there are no eigenvalues of A on this interval?

    Also, ##f(x)## is continuous since it is an nth degree polynomial.
     
  10. Aug 6, 2017 #9

    fresh_42

    Staff: Mentor

    Yes, that's right. We have an ##x## with ##f(x)=0##. Your conclusion is basically correct. I only would try to demonstrate why this is a contradiction. Where does ##f(x)=0## lead to? Where do the eigenvalues come in here?
     
  11. Aug 6, 2017 #10
    I'm not completely sure where to go with ##\det (A - xI)= 0##. Of course, the x's for which this is true are the eigenvalues of A, but I don't see how this is related to the interval [-1,1]
     
  12. Aug 6, 2017 #11

    fresh_42

    Staff: Mentor

    What does it mean, if a linear function has zero determinant?
     
  13. Aug 6, 2017 #12
    That it's not invertible?
     
  14. Aug 6, 2017 #13

    fresh_42

    Staff: Mentor

    Yes, and thus it has a non-trivial kernel, i.e. a vector ##v## with ##(A-x\cdot 1)(v)=0##. Now you can use ##A^3=A+1## again and apply it on ##v##.
     
  15. Aug 6, 2017 #14
    All I get is that ##x## must satisfy the equation ##x^3= x+1##
     
  16. Aug 6, 2017 #15

    Ray Vickson

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    Eigenvalues work perfectly well here. Since ##A^3 - A - I = 0##, ##p(x) = x^3 - x - 1## is the "minimal polynomial" of ##A##, and that means that its roots are all the eigenvalues of ##A##, not counting multiplicity. One of the eigenvalues is ##\lambda_1 > 0##, as we can see from the fact that ##p(x)## changes sign between ##x = 0## and large positive ##x##. Setting ##p'(x) = 0## implies that there are two stationary points, ##x = \pm 1/\sqrt{3}##, and the second-derivative test shows that ##x = -1/\sqrt{3}## is a (local) maximum, while ##x = +1/\sqrt{3}## is a local minimum. Therefore, there can be only one positive root of ##p(x) = 0##. Also, ##p(-1/\sqrt{3}) < 0##, so ##p(x)## never rises above the 0 for ##x < 0##, hence there are no negative real roots. Therefore, the other two eigenvalues are complex conjugates ##\lambda_2## and ##\lambda_3 = \overline{\lambda}_2##, hence ##\lambda_2 \times \lambda_3 > 0##. The characteristic polynomial ##c(x)## of ##A## must be a multiple of ##p(x)## but with the same roots, hence ##c(x) = (x-\lambda_1)^r (x - \lambda_2)^s (x - \lambda_3)^t##. The matrix ##A## is real, so ##c(x)## is a real polynomial and so its complex roots must come in conjugate pairs. This means that we must have have ##s = t \geq 1##. Thus, we can, indeed, determine the sign of the product of the eigenvalues.
     
    Last edited: Aug 7, 2017
  17. Aug 6, 2017 #16

    fresh_42

    Staff: Mentor

    Yes, and this is a) the solution and b) as I said, the eigenvalue argument by foot. You get that for some real number ##x_0## between ##x=-1## and ##x=1## that it has to hold ##x_0^3-x_0-1 = 0##. Until here you have used, that the function ##f(x)## is continuous, real-valued, that
    and that ##v \neq 0##.

    All which is left is to calculate the real number ##x_0## or at least that it has to be greater than ##1##.
     
  18. Aug 6, 2017 #17
    To calculate that real number, is it okay and necessary to use software?
     
  19. Aug 6, 2017 #18

    fresh_42

    Staff: Mentor

    That's not me to decide it. I also looked it up on the Wolfram page. O.k., I had ##A+x\cdot I## and ##x_0 = -1.3...## but this doesn't make any difference. I added (basically) "or at least outside of ##[-1,1]##" to my last post. This way one could do it analytically: calculate inflection points or show that there cannot be a zero between ##-1## and ##1## or even use Cardano's formulas. It would of course make some more work than the software solution, but it's possible.

    If we have the polynomial ##p(x)=x^3-x-1## then ##p(-1)=-1## and ##p(1)=-1##. Thus if it crossed the ##x-##axis in between, it would have to have a maximum ##x_m## anywhere in between with ##p(x_m)>0##. But this can easily be calculated by differentiation, solving a quadratic formula and evaluation of ##p(x_m)##. So this is one possible way without software.
     
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