- #1
kehler
- 104
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Does linear algebra go in this thread??
Anyway,
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)
dim V = dim I am + dim ker
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 I am = 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 :)
Anyway,
Homework Statement
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)
Homework Equations
dim V = dim I am + dim ker
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 I am = 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 :)