# Linear Transformation help !!

• February 2nd 2013, 10:31 AM
dave52
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?
• February 2nd 2013, 01:22 PM
jakncoke
Re: Linear Transformation help !!
let me just be clear. $[T]_{uv}$ is the transformation that goes from Vector Space U to V ?
• February 2nd 2013, 01:31 PM
dave52
Re: Linear Transformation help !!
u is a basis for M(2x2) and v is a basis for P(3)
• February 2nd 2013, 02:04 PM
jakncoke
Re: Linear Transformation help !!
so since any vector in $M_{2x2}(\mathbb{R})$ can be written as a linear combination of basis vectors $\{b_1,b_2,b_3,b_4\}$
$c_1b_1 + c_2b_2 + c_3b_3 + c_4b_4$ is any vector in V. since $L(c_1b_1 + c_2b_2 + c_3b_3 + c_4b_4) = c_1L(b_1)+...+c_4L(b_4)$ we need to find out the coordinates of the linear transformation of the basis vectors of V with respect to basis of U.

let $b_1 = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$, $b_2 = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$ , $b_3 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, $b_4 = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$
so $L(b_1) = 1 + x - x^2 + x^3$, which written in terms of the basis of U is $\begin{bmatrix} x & x-x^2 & x-x^3 & x-1 \end{bmatrix}$* $\begin{bmatrix} 2 \\ 1 \\ -1 \\ -1 \end{bmatrix} = 2x + x - x^2 -x + x^3 -x + 1 = x - x^2 + x^3 + 1 = L(b_1)$. I did this for all basis vectors of V.
and i got the following Linear Transformation Matrix from V to U

$\begin{bmatrix}2 & -2 & 0 & -2 \\ 1 & -1 & 0 & 1 \\ -1 & 1 & 1 & 0 \\ -1 & 1 & -1 & 0 \end{bmatrix}$ which has only 3 linearly indepdent vectors.
So $Dim(Img(T)) = 3$ so Dim(Ker(T)) = 1. Which is as so since only the matrix consisting of all entries identical $\begin{bmatrix}a & a \\ a & a \end{bmatrix} a \in \mathbb{R}$ 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 $\begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix}$

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

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.