<s>[INST] Does there exist a linear transformation from $\mathbb{R}^3 \to\mathbb{R}^3$ such that $T((1,1,1)) = (1,2,3), T((1,2,1) = (1,1,1)$? I tried making the column vectors $(1,1,1)$ and $(1,2,1)$ into a matrix, but because it's $3\times2$, it's noninvertible. Does this mean that I cannot recover the linear transformation without a third vector? Do the vectors that define the transformation $(1,1,1)$ and $(1,2,1)$ have to form a basis for the subspace they're being transformed from for the transformation to be defined? Sorry if I'm mixing things up or wording things in a confusing manner.
 [/INST] You're looking for a matrix  $ \begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}$ such that $$\begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j 
\end{pmatrix}
\begin{pmatrix}
1\\
1\\
1 
\end{pmatrix}
= \begin{pmatrix}
1\\
2\\
3 
\end{pmatrix}
$$
and
$$\begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j 
\end{pmatrix}
\begin{pmatrix}
1\\
2\\
1 
\end{pmatrix}
= \begin{pmatrix}
1\\
1\\
1 
\end{pmatrix}
.$$ If you do the multiplications you get six equations 
$$\begin{cases}
a+b+c=1 \\
e+f+g=2 \\
h+i+j=3 \\
a+2b+c=1 \\
e+2f+g= 1 \\
h+2i+j=1
\end{cases}.$$ You can easily find (by subtraction [e.g. $a+b+c=1$ and $a+2b+c=1$]) that $b=0$, $f=-1$, $i=-2$. Then for example choose $a=1, c=0, e=3,g=0, h=5,j=0$ and you can conclude that $$  \begin{pmatrix}
1&0&0\\
3&-1&0\\
5&-2&0
\end{pmatrix}$$ is a solution to your problem.
</s>