<s>[INST] Is there a linear operation such that $F(1,1,1) = (1,2,3),F(1,2,3) = (1,4,9),F(2,3,4) = (1,8,27)$? The exercise asks me verify if there exists a linear operator $F$ such that:
$$F(1,1,1) = (1,2,3)\\F(1,2,3) = (1,4,9)\\F(2,3,4) = (1,8,27)$$
First I tried to write a vector $(x,y,z)$ as a linear combination of $(1,1,1),(1,2,3),(2,3,4)$:
$$(x,y,z) = a(1,1,1) + b(1,4,9) + c(1,8,27)$$
then we have:
$$x = a + b + 2c\\y = a + 2b + 3c\\z = a + 3b + 4c$$
which is a system that has determinant $0$, therefore there isn't a way to represent a vector $(x,y,z)$ as a linear combination of $(1,1,1),(1,2,3),(2,3,4)$.
If it were possible to write the vector $(x,y,z)$ I would then apply $F$ to both sides and find the linear operator. But I can't do it since that vectors do not form a basis. But my reasoning does not logically prove that such linear operator does not exsists. 
By the way, my book also says that this is impossible because $(2,3,4)$ should be equal to $(2,6,12)$. Why?
 [/INST] The vectors $v_1=(1,1,1)$, $v_2=(1,2,3)$ and $v_3=(2,3,4)$ are not linearly independent; Gaussian elimination gives
\begin{align}
\begin{bmatrix}
1 & 1 & 2 \\
1 & 2 & 3 \\
1 & 3 & 4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 2 & 2
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\end{align}
which says that $v_3=v_1+v_2$.
Since $(1,2,3)+(1,4,9)=(2,6,12)\ne(1,8,27)$ no linear map exists with the requested property.
Had the problem said $F(2,3,4)=(2,6,12)$, then infinitely many linear maps would exist: complete $\{(1,1,1),(1,2,3)\}$ to a basis and send the third basis vector to any vector.

In the RREF above, the dominant columns (those with a leading $1$) are the first and the second column. The coefficients in the third column give exactly the coefficients to use for getting it as a linear combination of the dominant columns to its left.
Change the third column to represent the vector $(1,0,-1)=2v_1-v_2$; Gaussian elimination is
\begin{align}
\begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 0 \\
1 & 3 & -1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 & -1 \\
0 & 2 & -2
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix}
\end{align}
and indeed we find the right coefficients.
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