let me just be clear. is the transformation that goes from Vector Space U to V ?
Let T: M_{2x2}(R)------>P_{3}(R) be a linear transformation defined by
T ([a b]) = (a-b) + (a-d)x +(b-c)x^{2} +(c-d)x^{3}
[c d]
Consider the bases v ={[1 0] , [0 1] ,[1 0] ,[ 0 0]} of M_{2x2}
- - - -------- -- -- ---- [1 0 ] [0 1] [0 1] [1 1]
and u = {x, x-x^{2},x-x^{3},x-1} of P_{3}(R)
a) find [T]_{uv }
b) use [T]_{uv} to find a basis for the kernel of T
c) use [T]_{uv} to find a basis for the image of T
d) State the nullity and rank of T. Is T injective? surjective?
so since any vector in can be written as a linear combination of basis vectors
is any vector in V. since we need to find out the coordinates of the linear transformation of the basis vectors of V with respect to basis of U.
let , , ,
so , which written in terms of the basis of U is * . I did this for all basis vectors of V.
and i got the following Linear Transformation Matrix from V to U
which has only 3 linearly indepdent vectors.
So so Dim(Ker(T)) = 1. Which is as so since only the matrix consisting of all entries identical maps to the 0 matrix under this Linear Transformation.
Nullity is the Dim of the Kernel which is 1
Rank if the Dim of the image which is 3.
So the basis for the kernel is just
The basis for the image of the Linear Transformation are the following coordinates in P
Since the Ker contains more than just the 0 vector, it is certainly not injective. Since the Rank of T is 3 (Dim of codomain is 4), it is also not surjective.