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Linear independency of operators

  1. Mar 27, 2007 #1


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    OK, I stumbled upon a problem, but I feel somehow stupid about writing the exact problem down, so I'll ask a more "general" question.

    I have to see if three linear operators A, B and C from the vector space of all linear operators from R^2 to R^3 are linearly independent. The mappings are all known (i.e. A(x, y) = ... , etc.).

    Well, I set up an equation aA + bB + cC = 0, or, more precise, (aA + bB + cC)(x, y) = 0(x, y), where a, b and c are scalars. This equation should hold for all ordered pairs (x, y) from R^2 in order for these operators to be equal. Now, this equation led to a system of three equations with three unknowns, a, b and c (of course, the coefficients of the system are terms involving linear combinations of x and y). Now, since this must hold for every (x, y) from R^2, my logic was to plug in any (x, y) and solve for a, b and c. If the solution is trivial, then the operators are independent.

    Nevertheless, there is obviously something wrong with my logic, since, after plugging in, for example (x, y) = (0, 1), I obtain a solution in parametric form, i.e. there is no unique trivial solution!

    Any help is appreciated.
  2. jcsd
  3. Mar 27, 2007 #2


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    Suppose (a',b',c') is a solution to (aB + bB + cC)(0, 1) = (0, 0, 0). Then so is (ta',tb',tc'); observe:

    ((ta')A + (tb')B + (tc')C)(0,1)
    = (t(a'A) + t(b'B) + t(c'C))(0,1)
    = t(a'A + b'B + c'C)(0,1)
    = t(0,0,0) ... [since (a',b',c') is a sol'n]
    = (0,0,0)

    So in your situation, you've come up with the solutions being (ta',tb',tc') where t is a parameter. You found this after setting (x,y) = (0,1). This tells you that IF there is are non-trivial solutions to the equation aA + bB + cC = 0, they are of the form (ta',tb',tc') for some non-zero t. And it's clear that these are all solutions iff (a',b',c') is a solution. So compute:

    (a'A + b'B + c'C)(1,0)

    if you get (0,0,0), then you've found your non-trivial solution(s). If not, only the trivial solution exists.
  4. Mar 28, 2007 #3


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    AKG, thank you for your reply.

    I seem to have "temporarily forgotten" that the set of all solutions to a homogenous system is a vector space. :rolleyes:

    I did as you suggested, and I found out that the operators are dependent, since I arrived at (a'A + b'B + c'C)(1,0) = (0, 0, 0).
  5. Mar 29, 2007 #4


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    Well, a new dilemma is here.

    I found an exercise which is similar and solved. It's, again, about "testing" the independency of three linear functionals (where all of them are well known and all are from R^3 --> R).

    Anyway, at a point one obtains a(x1 - 2x2 + x3) + b(x1 + x2 + x3) + c(x1 - x2 - x3) = 0, for some scalars a, b, c, and this equation should hold for all x1, x2, x3 from R (of course, f1(x1, x2, x3) = x1 -2x2 + x3, etc.). Now, the solution says that we have to plug in (1, 0, 0), (0, 1, 0) and (0, 0, 1) into the equation in order to obtain a system of 3 equations with 3 unknowns which has a trivial solution a=b=c=0, so the functionals are independent.

    This looks perfectly logical, but it doesn't seem to be in consistency with the upper posts (unless I'm missing something huge). Can it perhaps be that, if the equations hold for all the basis vectors from R^3, then they hold for all vectors from R^3, too? What is the exact reason why the solution looks like I said it looks? Could we have plugged in any other three linearly independent vectors in order to obtain a 3x3 system whose solution tells us about the dependency/independency of the given functionals?

    Thanks in advance.

    Edit: I'm sorry the thread looks so homework-ish, since this isn't a homework section.
    Last edited: Mar 29, 2007
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