Answer to Question #254834 in Linear Algebra for Sabelo Xulu

Question

3. In each of the following cases explain whether R²$\to$ R is a linear transformation, if it is, supply a proof, if not, supply a counter ample

a) T(a, b) =a + b

b) T(a, b) =ab

c) T(a, b) =|a|²

d) T(a, b) =a - b

Answer 1

$\begin{aligned}T: \mathbb{R^2} \rightarrow \mathbb{R}\\ (a)\ T(a, b)=a+b & \\ T\left(\left(a_{1}, b_{1}\right)+\left(a_{2}, b_{2}\right)\right) &=T\left(a_{1}+a_{2}, b_{1}+b_{2}\right) \\ &=a_{1}+a_{2}+b_{1}+b_{2} \\ &=a_{1}+b_{1}+a_{2}+b_{2} \\ &=T\left(a_{1}, b_{1}\right)+T\left(a_{2}, b_{2}\right) \end{aligned}$

$\begin{aligned} T(\alpha(a, b))=& T(\alpha a, \alpha b) \\ &=\alpha a+\alpha b=\alpha(a+b)=\alpha T(a, b) \end{aligned}$

Thus $T$ is linear transformation.

$(b) \quad T(a, b)=a b\\ \text{ This is not linear transformation}$

$\begin{aligned} &\text { as } T(1,1)=1, T(1,0)+T(0,1)=0+0=0 \\ &\Rightarrow T(1,1) \neq T(1,0)+T(0,1) \end{aligned}$

$(c)\ T(a, b)=|a|^{b}\\ \text{This is not linear transformation.}$

$\begin{aligned} &T(2,1)=2^{1}=2 \\ &T(2,0)=2^{0}=1 \quad \& \quad T(0,1)=0^{1}=0 \\ &T(2,0)+T(0,1)=1 \neq T(2,1) \end{aligned}$

(d) $T(a, b)=a-b$

$\begin{aligned} & \\ T\left(\alpha\left(a_{1}, b_{1}\right)+\left(a_{2}, b_{2}\right)\right) &=T\left(\alpha a_{1}+a_{2}, \alpha b_{1}+b_{2}\right) \\ &=\alpha a_{1}+a_{2}-\alpha b_{1}-b_{2} \\ &=\alpha\left(a_{1}-b_{1}\right)+a_{2}-b_{2} \\ &=\alpha T\left(a_{1}, b_{1}\right)+T\left(a_{2}, b_{2}\right) \end{aligned}$

Thus, T is linear transformation.