My question is about linear transformations:
Is there a unique linear transformation $T:R^3\to R^3$ such that the image of the plane $M=\{t(1,4,0)+s(1,1,1)+(2,2,1)|t,s$$ \in$$R \}$ is the point $(0,3,1)$?
The question is not easy for me to answer, The 3 vectors $(1,4,0),(1,1,1),(2,2,1)$ are linearly independent, so it gives that $T$ is a unique (linear?) transformation (since $T$ maps 3 independent vectors in $R^3$), But on the other side $T(M)$ should give a result involved with $t,s$ (since for a linear transformation: $T(t \cdot \overrightarrow x) = t \cdot T(\overrightarrow x))$, what (I think) makes $T$ non linear transformation.
Does $T(M) = (0,3,1)$ say that $T(t(1,4,0))=T(s(1,1,1))=T(2,2,1)=(0,3,1)$, or $T(t(1,4,0))+T(s(1,1,1))+T(2,2,1)=(0,3,1)$? Where the scalars t, s appear in the result?
Since $T(M) = (0,3,1)$, i think that for every $ \overrightarrow x \in M$, $T(\overrightarrow x) = (0,3,1)$. Please tell me if I am right or not.
Thanks in advance for help!
The three vectors $(1,4,0),(1,1,1),(2,2,1)$ are are linearly independent, hence there exists a unique linear transformation $T$ with
$T(2,2,1)=(0,3,1), T(1,4,0)=(0,0,0)$ and $T(1,1,1)=(0,0,0)$.
Therefore $T(M)=\{(0,3,1)\}.$
On the other hand, if $S$ is a linear transformation with $S(M)=\{(0,3,1)\}$, then we have , since $(2,2,1) \in M$ and $(2,2,1)+(1,4,0) \in M$,
$S((2,2,1)+(1,4,0))=(0,3,1)+S(1,4,0)=(0,3,1)$, thus $S(1,4,0)=(0,0,0)$.
In a similar way we see that $S(1,1,1,)=(0,0,0)$.
Consequence: $S=T$.