Linear Transformation help !!

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?

Re: Linear Transformation help !!

let me just be clear. $\displaystyle [T]_{uv} $ is the transformation that goes from Vector Space U to V ?

Re: Linear Transformation help !!

u is a basis for M(2x2) and v is a basis for P(3)

Re: Linear Transformation help !!

so since any vector in $\displaystyle M_{2x2}(\mathbb{R})$ can be written as a linear combination of basis vectors $\displaystyle \{b_1,b_2,b_3,b_4\} $

$\displaystyle c_1b_1 + c_2b_2 + c_3b_3 + c_4b_4 $ is any vector in V. since $\displaystyle L(c_1b_1 + c_2b_2 + c_3b_3 + c_4b_4) = c_1L(b_1)+...+c_4L(b_4) $ we need to find out the coordinates of the linear transformation of the basis vectors of V with respect to basis of U.

let $\displaystyle b_1 = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} $, $\displaystyle b_2 = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} $ ,$\displaystyle b_3 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $, $\displaystyle b_4 = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} $

so $\displaystyle L(b_1) = 1 + x - x^2 + x^3 $, which written in terms of the basis of U is $\displaystyle \begin{bmatrix} x & x-x^2 & x-x^3 & x-1 \end{bmatrix}$* $\displaystyle \begin{bmatrix} 2 \\ 1 \\ -1 \\ -1 \end{bmatrix} = 2x + x - x^2 -x + x^3 -x + 1 = x - x^2 + x^3 + 1 = L(b_1) $. I did this for all basis vectors of V.

and i got the following Linear Transformation Matrix from V to U

$\displaystyle \begin{bmatrix}2 & -2 & 0 & -2 \\ 1 & -1 & 0 & 1 \\ -1 & 1 & 1 & 0 \\ -1 & 1 & -1 & 0 \end{bmatrix} $ which has only 3 linearly indepdent vectors.

So $\displaystyle Dim(Img(T)) = 3 $ so Dim(Ker(T)) = 1. Which is as so since only the matrix consisting of all entries identical $\displaystyle \begin{bmatrix}a & a \\ a & a \end{bmatrix} a \in \mathbb{R} $ 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 $\displaystyle \begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix} $

The basis for the image of the Linear Transformation are the following coordinates in P $\displaystyle \begin{bmatrix}2 & 0 & -2 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \\ -1 & -1 & 0 \end{bmatrix} $

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.