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  • 2 years ago

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)

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  1. Jemurray3
    • 2 years ago
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    Do you have an idea of where to start?

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

  3. REMAINDER
    • 2 years ago
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    @phi

  4. Jemurray3
    • 2 years ago
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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.

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

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

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

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

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

  10. Jemurray3
    • 2 years ago
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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

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

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

  13. Jemurray3
    • 2 years ago
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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

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

  15. phi
    • 2 years ago
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    yes kernel is empty

  16. Jemurray3
    • 2 years ago
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    Yep

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

  18. Jemurray3
    • 2 years ago
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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.

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