Question

show that the function T: R2 → R2 such that T(0,1)=(3,4),T(3,1)=(2,2)

And T(3,2)=(5,7) is not a L.T.

Answer 1

Linear Transformation :
Let V and W be two vector spaces on K (k= Real / Complex) .

A map T: V ---> W is said to be linear transformation iff:
a) Transformation : T(x+y) = T(x) + T(y) : for all x, y belongs to V.
b) T(cx)= cT(x) , for all c belongs to scalar, x belongs to V.

Here:
a) T (0,1) = (3,4)
T(3,1) = (2,2 ) T
Then,
T( 0,1)+ T(3,1) = (3,4) +(2,2) = (5,6) is not equal to
T(3,2) = (5,7)
Hence, It doesn't satisfy condition 1 :
Therefore, This is not a Linear Transformation.

Answer 2

A linear transformation between two vector spaces $V$ and $W$ is a map $T:V\rightarrow W$ in such a way that ,
1. $T(v_1+v_2)=T(v_1)+T(v_2)$ for any vector $v_1$ and $v_2$.
2. $T(\alpha v) =\alpha T(v)$

Here , v₁ = (0,1) and v₂ = (3,1)
so, from condition (1),
T(v₁ + v₂) = T{(0,1) + (3,1) } = T(0,1) + T(3,1)
e.g., LHS = T{(0,1) + (3,1)} = T(3,2) , it is given (5,7)
so, LHS = (5,7)
Now, RHS = T(0,1) + T(3,1)
= (3,4) + (2,2) = (5,6)
Did you observed LHS ≠ RHS ? Of course LHS ≠ RHS
Hence, function $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that T(0,1) = (3,4) , T(3,1) = (2,2) and T(3,2) is not linear transformation.