Homework Help: Dimension, Image, Kernels etc - Linear Algebra stuff

1. Aug 23, 2008

kehler

Does linear algebra go in this thread??
Anyway,
1. The problem statement, all variables and given/known data
Let n be an integer at least 1, and x1,....,xn be distinct points in R. For any integer m>=1, let P denote the vector space of polynomials of degree at most m. Define a linear transformation T:P->R^n by
[f(x1)]
[f(x2)]
T(f)= [ . ] <--- (That's meant to be one big bracket)
[ . ]
[ . ]
[f(xn)]

What is the dimension of the kernel of T? What is the dimension of the range of T? (your answer will depend on m and n)

2. Relevant equations
dim V = dim Im + dim ker

3. The attempt at a solution
Well I've considered the case when n>m. In this case, dim V= m+1 and dim ker = 0 as it's not possible for any polynomial of degree m+1 to have more than m roots. So I find that dim Im = m+1.
Now for m>=n, I don't seem to be getting anywhere. Is the dimension of the kernel m-1 cos there can be m-1 roots?

Any help would be really appreciated :)

2. Aug 23, 2008

kehler

Argh it messed up my bracket.
Well T(f) is meant to be
[f(x1)]
[f(x2)]
[f(x3)] <- it's meant to be a single long bracket extending downwards
[......]
[......]
[......]
[f(xn)]

3. Aug 24, 2008

HallsofIvy

So T(f) maps f into (f(x1), f(x2), f(x3)...)?

Okay, f is in the kernel of T if and only if f(x1)= 0, f(x2)= 0, ...

(y1, y2, y3, ...) is in the image of T if and only if there exist a polynomial f such that f(x1)= y1) f(x2)= y2, ...

4. Aug 24, 2008

kehler

Yea that's right. So when m>n, the polynomial will have m roots. In this case is the dimension of the kernel 1 and the dimension of the image m? I'm kinda confuse about how I can tell what the dimension is.
(The dimension just means the number of vectors in the basis)

5. Aug 24, 2008

Defennder

I'd like to clarify something here. Does the transformation T operate only on (x1,x2...xn) where all the xi's are distinct? Suppose for eg. m=2, then f(x) has two roots and the vectors in ker(T) are of the form (x1,x2,x1,x1...) <-any permutation of x1 and x2, which grows to an exponentially large number.

6. Aug 24, 2008

HallsofIvy

The transformation T operates on polynomials. I'm going to assume that x1, x2, etc. are distinct.

For example, suppose $x_1= 0$, $x_2= 1$ and m= 2 so P is the vector space of polynomials of order at most 2.

Then T(p(x))= (p(0), p(1)). The kernel is the subspace of all such polynomials such that p(0)= 0, p(1)= 0. Since p(x)= a+ bx+ cx2, p(0)= a= 0 and p(1)= b+ c= 0. Since c= -b, we are free to choose b and then a= 0, c= -b so the kernel has dimension 1. Since P itself has dimension 3, it follows that the image of T has dimension 2.

More generally, P has dimension m+ 1. Setting P(x1)= 0, ..., P(xn)= 0, we have n equations for those m+ 1 variables. That should leave m+1- n independent variables so it looks to me like the kernel of T has dimension m-n+1 and the image has dimension m+1- (m-n+1)= n.

Of course, that's not a proof and may be completely wrong.

7. Aug 24, 2008

kehler

That seems correct :). But what if m+1<n? The kernel can't have a negative dimension, can it?

8. Aug 24, 2008

HallsofIvy

If that is the case, then we have more points than we have coefficients: but we will still have the trivial polynomial, p(x)= 0 for all x. The smallest the kernel can be is the 0 vector only and that has dimension 0. As long as $m+1\le n$, the kernel will have dimension 0 and the image will have dimension m+1.

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