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REMAINDER Group Title

suppose that T: P2 -> P3 is a linear transformation where T(1)=1+x^2 T(x)=x^2 -x^3 T(x^2)=2+x^3 obtain the bases for Im(T) and ker(T)

  • 2 years ago
  • 2 years ago

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  1. Jemurray3 Group Title
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    Do you have an idea of where to start?

    • 2 years ago
  2. REMAINDER Group Title
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    i have this but i'm not sure |dw:1349109417654:dw|

    • 2 years ago
  3. REMAINDER Group Title
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    @phi

    • 2 years ago
  4. Jemurray3 Group Title
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    Well your transformation matrix should be \[\left(\begin{matrix}1 & 0 & 2 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & -1 & 1 \end{matrix}\right)\] check to make sure you see why.

    • 2 years ago
  5. phi Group Title
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    did you start with Jem's matrix ?

    • 2 years ago
  6. REMAINDER Group Title
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    it does not sure it written math processing error

    • 2 years ago
  7. Jemurray3 Group Title
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    Works for us... it should be |dw:1349109968192:dw|

    • 2 years ago
  8. REMAINDER Group Title
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    |dw:1349110409182:dw| then wat would be Im(T) and ker(T) if it like this?

    • 2 years ago
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    Im(T)={+x^2;x^2-x^3;2+x^3} is it gonna be this?

    • 2 years ago
  10. Jemurray3 Group Title
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    The image is the set of all vectors y that you could get by applying the transformation T. The three columns of that matrix are linearly independent, which means that there is no nontrivial nullspace. However, because there are more rows than columns, the image cannot span all of P^4. Obviously the transformation will never yield a quadratic term because none of the columns have a nonzero entry in the second row. So the image would be something like all polynomials of degree <=4 of the form a + bx + cx^3

    • 2 years ago
  11. Jemurray3 Group Title
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    I'm sorry, of degree <= 3

    • 2 years ago
  12. REMAINDER Group Title
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    what u mean is the image will |dw:1349111483400:dw|

    • 2 years ago
  13. Jemurray3 Group Title
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    Oh I made a little mistake... it should be that there'll be no linear term, not quadratic. So the image will be the set of all polynomials of the form a + bx^2 + cx^3

    • 2 years ago
  14. REMAINDER Group Title
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    ok, then the kernel =0 or what?

    • 2 years ago
  15. phi Group Title
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    yes kernel is empty

    • 2 years ago
  16. Jemurray3 Group Title
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    Yep

    • 2 years ago
  17. REMAINDER Group Title
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    do i write zero or is the any way to represent empty for the kernel?

    • 2 years ago
  18. Jemurray3 Group Title
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    You could just say \[ker(T) = \vec{0} \] or you could write that it's empty, or only the zero vector, or whatever you want I suppose.

    • 2 years ago
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