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Linear Independence: Polynomial Example

  1. Oct 3, 2010 #1
    Hello everyone. I was going through my Linear Algebra Done Right textbook that threw me off. I hope this forum is appropriate for my inquiry; while there is no problem I'm trying to solve here, I don't know whether just asking for clarification would belong to the homework forum instead. If this is the case, I apologize.

    I'll start off by quoting the passage:

    "For another example of a linearly independent list, fix a non-negative integer m. Then (1,z,...,zm) is linearly independent in P(F). To verify this, suppose that a0, a1,...,am, belonging to F are such that:

    a0 + a1z + ... + amzm = 0, for every z belonging in F.

    If at least one of the coefficients a0, a1,...,am were nonzero, then the above equation could be satisfied by at most m distinct values of z; this contradiction shows that all the coefficients in the above equation equal 0. Hence (1,z,...,zm) is linearly independent, as claimed."

    Linear independence, as I understand it, holds only when each vector in a list of vectors has a unique representation as a linear combination of other vectors within that list. It is my interpretation that Axler is specifically using the fact that the {0} vector, in the above polynomial vector space example, can only be expressed by setting all coefficients to 0.

    My confusion, I think, stems from how he concludes that all the coefficients must be zero. If any coefficient is nonzero, then the equation has, at most, m roots (I hope I am correctly relating this to the Fundamental Theorem of Algebra). But then, as I see it, this shows that the equation has more than one representation for {0} and is thus not linearly independent. But instead, he uses this same fact to obtain a contradiction and conclude that all the coefficients must equal 0.

    Unfortunately, for some reason, the TeX editor did not work properly for me, so I had to resort to expressing some things here differently. Anyway, if anyone could shed some light and point me in the right direction, I would greatly appreciate it.
  2. jcsd
  3. Oct 3, 2010 #2


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    There is your confusion. Yes, there may exist many values of z which make the numerical value of the function 0. But that is NOT what is being discussed here. To say that the polynomial [itex]a_0+ a_1z+ a_2z^2+ \cdot\cdot\cdot+ a_nz^n= 0[/itex] means that it is 0 for all z, not just for some.

    If the polynomial [itex]a_0+ a_1z+ a_2z^2+ \cdot\cdot\cdot+ a_nz^n= 0[/itex] for all z, then, in particular, that is true for z= 0. Setting z= 0 in that gives [itex]a_0= 0[/itex]. But if it is 0 for all z, then it is a constant- its derivative is 0 for all z. It's derivative is [itex]a_1+ 2a_2z^1+ \cdot\cdot\cdot+ na_nz^{n-1}= 0[/itex] for all z. Setting z= 0 that gives [itex]a_1= 0[/itex]. But since that is 0 for all z, it is a constant and its derivative must be 0 for all z. We can proceed by induction to show that if a polynomial in z is 0 for all z, then all of its coefficients must be 0.
  4. Oct 3, 2010 #3
    Ah, that makes it all clear now. Thanks!
  5. Oct 4, 2010 #4
    (Slightly pedantic here: The statement that for a polynomial p over F, p = 0 if and only if p(a) = 0 for all a in F is only true if F is infinite. If F is finite, with elements {a1,...,aq}, then the polynomial (x - a1)...(x - aq) is nonzero but evaluates to zero at every element in F. Remember that a polynomial is defined by its coefficients; a polynomial is zero if and only if all its coefficients are zero.)
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