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Determinant of functions

  1. Jul 31, 2013 #1
    1. The problem statement, all variables and given/known data
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    3. The attempt at a solution

    I have already proved that ## N(T) \supset span\{y_{1},y_{2},...y_{n} \} ## but am having trouble proving the other half. You can't use the fact that the determinant is 0 to conclude that functions are linearly dependent unless you also know that the functions are analytic. Any tips on getting around this?

    Thanks!

    BiP
     
    Last edited by a moderator: May 6, 2017
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  3. Jul 31, 2013 #2

    micromass

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    What does determinant being zero tell you about the rank of the matrix? What are some equivalent characterizations of the rank?
     
  4. Jul 31, 2013 #3
    I am a bit confused. When I think of rank of a matrix I generally think of the rank of a matrix with entries from a field. Here, the matrix entries are functions. Are my functions the elements of the field I am concerned with? Or is it that I am evaluating all the functions to form a matrix with entries from a field whose determinant is being computed to 0?

    So basically, am I correct in understanding that if y is in N(T), then basically what this means is that if you evaluate the matrix at any value of t, its determinant will always be 0. So for any value of t, the matrix will fail to have full rank since its determinant is always 0.

    Please clarify micro, thanks very much!!

    BiP
     
    Last edited: Jul 31, 2013
  5. Jul 31, 2013 #4

    HallsofIvy

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    If y is in the span of [itex]\{y_1, y_2, \cdot\cdot\cdot, y_n\}[/itex], then it can be written as a linear combination of vectors in that set. And that means that the first row is a linear combination of the succeeding rows and so the determinant is 0 for all t. And, the other way works. If y is in N(T), the "null space of T", that means, by definition of "null space", that T(y)= 0. That is, that this determinant is 0 for all t. From that it follows that (since we are given that all of [itex]y_1[/itex], [itex]y_2[/itex], ..., [itex]y_n[/itex] are independent) that the first row is a linear combination of the other rows and so y is a linear combination of the other vectors. That is, that y is in their span.
     
  6. Aug 1, 2013 #5
    Thanks! One thing still bothers me though, how does the determinant being 0 allow us to conclude that the first row is a LC of the other rows? This seems incorrect, since you cannot use the 0 Wronskian to conclude that a set of functions is linearly dependent, unless you also know that the functions are analytic?

    BiP
     
  7. Aug 1, 2013 #6

    HallsofIvy

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    The fact that the determinant is 0, alone, does NOT tell us that the first row is a linear combination of the other rows and I did not say it did. But since the determinant is 0, there does exist a "linear combination of rows" that is equal to 0 (the rows, thought of as vectors, are not linearly independent). Then, because we are told that the [itex]y_i[/itex] are independent, all except the first rows are independent.
     
  8. Aug 1, 2013 #7
    I'm not saying the statement is false, but somewhere I think your reasoning is flawed. I think you need to eventually utilize the fact that the functions are in ## C^{∞} ##. Otherwise I can cook up a counterexample:

    Suppose that you have the singleton set consisting of the function ## y_{1} = t^{2} ##. Now suppose that ## T(y) = 0 ##. Then if the theorem were true, it must be the case that ## y \in span(t^{2}) ##. But if we take ## y = t|t| ##, then it turns out that indeed ## T(y) = 0 ## but ## y \not \in span(t^{2}) ##.

    Of course, a problem with my counterexample is that the function I gave is not in ## C^{∞} ##. But certainly then, if you were to claim the theorem is true, you must use the fact that the functions ## y_{1},y_{2}...y_{n} ## are all linearly independent functions in ## C^{∞} ## rather than only using the fact that their determinant is 0!

    I'd like to see your opinion on this. Thanks!

    BiP
     
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