# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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.