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Showing that a linear transformation from P3 to R4 is an isomorphism?

  1. May 4, 2014 #1

    elo

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    I have a linear transformation, T, from P3 (polynomials of degree ≤ 3) to R4 (4-dimensional real number space). I have a second linear transformation, U, from R4 back to P3.

    In the first step of this four-step problem, I have shown that the composition TU from R4 to R4 is the identity linear transformation by sending a vector x through U, then sending the product of that through T, and ending up with x again.

    What I don't understand is the next step: show that T is an isomorphism. The problem gives the hint that I should first show that the image of T spans R4: okay, fair enough. After some misguided attempts to prove linear independence of the entries of the column vector of T(f), I skipped ahead to the solutions, and was given this:

    "Given a vector x in R4, we have T(U(x)) = x, by the previous part, and so x is a value of T, and im(T) = R4."

    I do not understand this answer. I did T(U(x)) = x, so I get that. But what does it mean that "x is a value of T?" And how does that mean that the image of T spans R4?
     
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  3. May 4, 2014 #2

    jbunniii

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    If you have already shown that ##TU=I##, then ##T## is surjective: if not, then the range of ##T## would not be all of ##\mathbb{R}^4##, and so the range of ##TU## could not be either, contradicting ##TU=I##. Can you use the rank-nullity theorem to conclude that ##T## is also injective?
     
  4. May 4, 2014 #3

    jbunniii

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    I think it's a typo or misstatement. It would be correct if it said "and so x is an element of im(T)".
     
  5. May 4, 2014 #4

    elo

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    Okay: since I can get a 4-dimensional x out of T, the image of T must span R4. The starting space P3 has four dimensions. By rank-nullity, the dimension of the kernel of the transformation plus the dimension of its image must equal the dimension of P3; since the dimension of the image is four, the dimension of the kernel must be zero. Thus, T is both surjective and injective, and is thus an isomorphism.

    Do I have the right idea?
     
  6. May 5, 2014 #5

    jbunniii

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    Yes, it sounds right to me!
     
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