## 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) one year ago one year ago

1. Jemurray3 Group Title

Do you have an idea of where to start?

2. REMAINDER Group Title

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

3. REMAINDER Group Title

@phi

4. Jemurray3 Group Title

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 Group Title

6. REMAINDER Group Title

it does not sure it written math processing error

7. Jemurray3 Group Title

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

8. REMAINDER Group Title

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

9. REMAINDER Group Title

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

10. Jemurray3 Group Title

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 Group Title

I'm sorry, of degree <= 3

12. REMAINDER Group Title

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

13. Jemurray3 Group Title

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 Group Title

ok, then the kernel =0 or what?

15. phi Group Title

yes kernel is empty

16. Jemurray3 Group Title

Yep

17. REMAINDER Group Title

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

18. Jemurray3 Group Title

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.