Question #254834

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



1
Expert's answer
2021-11-09T16:52:07-0500

T:R2R(a) T(a,b)=a+bT((a1,b1)+(a2,b2))=T(a1+a2,b1+b2)=a1+a2+b1+b2=a1+b1+a2+b2=T(a1,b1)+T(a2,b2)\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}

T(α(a,b))=T(αa,αb)=αa+αb=α(a+b)=αT(a,b)\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 TT is linear transformation.


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

 as T(1,1)=1,T(1,0)+T(0,1)=0+0=0T(1,1)T(1,0)+T(0,1)\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)=abThis is not linear transformation.(c)\ T(a, b)=|a|^{b}\\ \text{This is not linear transformation.}

T(2,1)=21=2T(2,0)=20=1&T(0,1)=01=0T(2,0)+T(0,1)=1T(2,1)\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)=abT(a, b)=a-b

T(α(a1,b1)+(a2,b2))=T(αa1+a2,αb1+b2)=αa1+a2αb1b2=α(a1b1)+a2b2=αT(a1,b1)+T(a2,b2)\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.


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