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

A linear transformation between two vector spaces $V$ and $W$ is a map $T:V\rightarrow W$ in such a way that ,
1. $T(v_1+v_2)=T(v_1)+T(v_2)$ where v₁ and v₂ are two vectors in space.
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