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anonymous
 3 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)
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you have an idea of where to start?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have this but i'm not sure dw:1349109417654:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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.

phi
 3 years ago
Best ResponseYou've already chosen the best response.0did you start with Jem's matrix ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it does not sure it written math processing error

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Works for us... it should be dw:1349109968192:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349110409182:dw then wat would be Im(T) and ker(T) if it like this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im(T)={+x^2;x^2x^3;2+x^3} is it gonna be this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, of degree <= 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what u mean is the image will dw:1349111483400:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, then the kernel =0 or what?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do i write zero or is the any way to represent empty for the kernel?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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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