# Linear Algebra Proof Regarding Linear Transformations

• Feb 28th 2013, 11:51 PM
zachoon
Linear Algebra Proof Regarding Linear Transformations
Given the following vectors in R^2:
a(sub 1) = (1; -1), a(sub 2) = (2; -1), a(sub 3) = (-3;2),
b(sub 1) = (1;0), b(sub 2) = (0;1), b(sub 3) = (1;1),
determine whether there is a linear transformation T from R^2 into R^2 such that Ta(sub i) = b(sub i) for i = 1, 2 and 3.

Basically, I don't quite understand the process of figuring out linear transformations, so I'd like at least a few helpful pointers in figuring this out. Thank you.
• Mar 1st 2013, 12:53 AM
jakncoke
Re: Linear Algebra Proof Regarding Linear Transformations
so assume we have a linear transformation. L which takes $\displaystyle a_i \to b_i$
$\displaystyle A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 2 \end{bmatrix}$
$\displaystyle B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$
For the matrix A, you can write the 3rd column as the linear combination of the first and second columns.
$\displaystyle A = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}$
since a linear transformation has the property L($\displaystyle c_1 v_1 + c_2 v_2$)= $\displaystyle c_1 L(v_1) + c_2 L(v_2)$ then
if the first column of A went to the first column of B, and the second column of A went to the second column of B then
$\displaystyle L(-1 * \begin{bmatrix} 1 \\ -1 \end{bmatrix} + -1 * \begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1*L(\begin{bmatrix} 1 \\ -1 \end{bmatrix}) + -1 * L(\begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1 * \begin{bmatrix} 1 \\ 0 \end{bmatrix} + -1 * \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$

since this is not equal to the 3rd column of B, there is no Linear transformation which takes the first column of A to the first column of B, second ....