## REMAINDER 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)

1. Jemurray3

Do you have an idea of where to start?

2. REMAINDER

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

3. REMAINDER

@phi

4. Jemurray3

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

6. REMAINDER

it does not sure it written math processing error

7. Jemurray3

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

8. REMAINDER

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

9. REMAINDER

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

10. Jemurray3

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

I'm sorry, of degree <= 3

12. REMAINDER

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

13. Jemurray3

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

ok, then the kernel =0 or what?

15. phi

yes kernel is empty

16. Jemurray3

Yep

17. REMAINDER

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

18. Jemurray3

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.