Question

Given a transformation T:R^2→R^2 defined as T(x_1,x_2 )=(0,x_1-x_2). Find ker⁡(T) and R(T), range of T

Answer 1

$$T\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix}\to\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$$

$$x_1=x_2$$

$$\ker(T)=\{(x_1, x_2)\in\R^2|x_1=x_2\}$$

Hence $\{(1,1)\}$ is a basis of $\ker(T).$

The set $\{(1,0)\}$ forms a basis for the range of $T, R(T).$

Rank theorem

$$rank(T ) + nullity(T ) = dim(\R^2)$$

$$1+1=2$$