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

- jamiebookeater

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- anonymous

Do you have an idea of where to start?

- anonymous

i have this but i'm not sure
|dw:1349109417654:dw|

- anonymous

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- anonymous

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.

- phi

did you start with Jem's matrix ?

- anonymous

it does not sure it written math processing error

- anonymous

Works for us... it should be |dw:1349109968192:dw|

- anonymous

|dw:1349110409182:dw|
then wat would be Im(T) and ker(T) if it like this?

- anonymous

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

- anonymous

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

- anonymous

I'm sorry, of degree <= 3

- anonymous

what u mean is the image will
|dw:1349111483400:dw|

- anonymous

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

- anonymous

ok, then the kernel =0 or what?

- phi

yes kernel is empty

- anonymous

Yep

- anonymous

do i write zero or is the any way to represent empty for the kernel?

- anonymous

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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