# Thread: Linear Transformation help !!

1. ## Linear Transformation help !!

Let T: M2x2(R)------>P3(R) be a linear transformation defined by
T ([a b]) = (a-b) + (a-d)x +(b-c)x2 +(c-d)x3
[c d]

Consider the bases v ={[1 0] , [0 1] ,[1 0] ,[ 0 0]} of M2x2
- - - -------- -- -- ---- [1 0 ] [0 1] [0 1] [1 1]

and u = {x, x-x2,x-x3,x-1} of P3(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?

2. ## Re: Linear Transformation help !!

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

3. ## Re: Linear Transformation help !!

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

4. ## 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.