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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)
 one year ago
 one year 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)
 one year ago
 one year ago

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Jemurray3Best ResponseYou've already chosen the best response.1
Do you have an idea of where to start?
 one year ago

REMAINDERBest ResponseYou've already chosen the best response.0
i have this but i'm not sure dw:1349109417654:dw
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
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.
 one year ago

phiBest ResponseYou've already chosen the best response.0
did you start with Jem's matrix ?
 one year ago

REMAINDERBest ResponseYou've already chosen the best response.0
it does not sure it written math processing error
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
Works for us... it should be dw:1349109968192:dw
 one year ago

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

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

Jemurray3Best ResponseYou've already chosen the best response.1
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
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
I'm sorry, of degree <= 3
 one year ago

REMAINDERBest ResponseYou've already chosen the best response.0
what u mean is the image will dw:1349111483400:dw
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
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
 one year ago

REMAINDERBest ResponseYou've already chosen the best response.0
ok, then the kernel =0 or what?
 one year ago

REMAINDERBest ResponseYou've already chosen the best response.0
do i write zero or is the any way to represent empty for the kernel?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
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.
 one year ago
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