Suppose v,u,w are distinct points on one line in $R^3$. The line need not pass through the origin. Prove that {v, u, w} is linearly dependent. There is a solution to an exercise from a linear algebra textbook written below. I am new to writing proof and I would like to see if my proof is correct. Thank you.
Suppose v,u,w are distinct points on one line in $ℝ^3$. The line need not pass through the origin. Prove that {v, u, w} is linearly dependent.
PROOF:
We can write v,u,w as a $3$ x $3$ matrix $A$=[v,u,w]
v,u,w are distinct points on one line so the columns of $A$ do not span $ℝ^3$. Therefore, the corresponding linear transformation $T(x)=Ax$ cannot be onto.
But, there is a pivot position in every row of $A$ IFF $T(x)$ is onto. Therefore we have less than 3 pivot positions and so the RREF of $A$ must have free variables.
So, the homogenous system of equations $A$x$=0$ must have non trivial solutions. But, the columns of $A$ are linearly independent IFF the corresponding homogenous system has only the trivial solution.
Therefor, v,u,w are linearly dependent.

A straight line is given by the formula $\mathbf{r}(t)=\mathbf{a}+\mathbf{b}t$. So if $\mathbf{u},\mathbf{v},\mathbf{w}$ are on the line, \begin{align}
\mathbf{u}&=\mathbf{a}+\mathbf{b}t_1\\
\mathbf{v}&=\mathbf{a}+\mathbf{b}t_2\\
\mathbf{w}&=\mathbf{a}+\mathbf{b}t_3\end{align}
Clearly, $\mathbf{u},\mathbf{v},\mathbf{w}\in Span(\mathbf{a},\mathbf{b})$ which is a 2-dimensional space, hence the three vectors must be linearly dependent. (Theorem: Every spanning set has more/equal elements than every linearly independent set of vectors.) This can also be seen directly by eliminating $\mathbf{a}$ and $t$, \begin{align}\mathbf{u}-\mathbf{v}&=\mathbf{b}(t_1-t_2)\\
\mathbf{v}-\mathbf{w}&=\mathbf{b}(t_2-t_3)\end{align}
$$\therefore \frac{\mathbf{u}-\mathbf{v}}{t_1-t_2}-\frac{\mathbf{v}-\mathbf{w}}{t_2-t_3}=\mathbf{0}$$ ($t_1\ne t_2\ne t_3$)
