# Determine a basis for the image of F

1. May 16, 2014

### victoranderson

Hello I have a question about part (b)

Now I know that F(f)(x) = [$\frac{1}{2}a+\frac{5}{6}b+\frac{17}{12}c]+[-a-2b-4c]x+[b+\frac{7}{2}c]x^{2}+[\frac{-2}{3}b-\frac{7}{3}c]x^{3}$

so Kernel of F : F(f)(x) =0

=> $\frac{1}{2}a+\frac{5}{6}b+\frac{17}{12}c=0$
$-a-2b-4c=0$
$b+\frac{7}{2}c=0$
$\frac{-2}{3}b-\frac{7}{3}c=0$

which gives a=3c,b=-7c/2
so the basis for the kernel of F is (3,-7/2,1)

I think it is correct.

After that I am confusing about the basis for the image of F

When I convert the system of equations into matrix form and use wolframalpha to solve it
there are basis for column space and basis for row space
I do not know which one is the one I am looking for.
I think it should not be using matrix to determine the basis for the image of F because it should be used in the latter part.

Last edited: May 16, 2014
2. May 16, 2014

### vela

Staff Emeritus
The basis you found for the kernel of F is correct.

To find a basis for the image, you might start by finding a set of vectors that span the image of F and then find the largest subset that's linearly independent.

3. May 16, 2014

### victoranderson

so the basis for the image is {$\frac{1}{2},\frac{5}{6},\frac{17}{12}$} and {$-1,-2,-4$}?

Another question.
Since F: P2(R) -> P3(R), the dim(F) is 3 or 4?
After doing this question I think dim(F) = 3, since dim(ker(F)) = 1 and dim(Im(F))=2

4. May 16, 2014

### vela

Staff Emeritus
No. For one thing, the elements in the image of F are four-dimensional. Your vectors are three-dimensional.

I don't know what you mean by "dim(F)".

Last edited: May 16, 2014
5. May 16, 2014

### victoranderson

That is my question.
How do we determine the basis for the image and the dimensions?
If you say the image of F is four-dimensional, then basis for the image of F can only be
span {(1/2,-1,0,0),(5/6,-2,1,-2/3),(17/12,-4,7/2,-7/3)} = span {(1/2,-1,0,0),(5/6,-2,1,-2/3)}
Is it correct now?

Last time I thought dim(F) = dim(im(F))+ dim (ker(F))
Actually it is dim(P2(R)) = dim(im(F))+ dim (ker(F). I understand now.

6. May 16, 2014

### vela

Staff Emeritus
Sorry, I misspoke earlier. The image of F isn't four-dimensional; it's a subspace of a four-dimensional space. My point was that the vectors in Im(F) should have four coordinates, not three, as you seemed to have inferred.

If you rewrite what you wrote earlier for F(f)(x) slightly, you get

F(f)(x) = a(1/2 + x) + b(5/6 - 2x + x2 - 2/3 x3) + c(17/12 - 4x + 7/2 x2 - 7/3 x3),

which you can interpret as a linear combination of three vectors. Hopefully, that's what you did to come up with the vectors above.

The vectors (1/2,-1,0,0) and (5/6,-2,1,-2/3) are linearly independent and span the image, so they are a basis for Im(F). There are two vectors, so the dimension of the image is 2.

Right. When you talk about dimensions, you're talking about the dimension of a vector space (or subspace). To ask what dim(F) is doesn't make sense because F is a mapping, not a vector space, and as you noted, the domain and codomain have different dimensions.

The relation dim(P2(R)) = dim(im(F))+ dim(ker(F)) reflects the fact that you can split the domain into two pieces. The vectors in one piece, ker(F), all map to 0. The vectors in the other piece map onto the image of F.