I'm independently studying Boyd & Vandenberghe's Convex Optimization and came across the following statement.

Suppose $x_1 \ne x_2$ are two points on $\mathbb{R}^n$. Points of the form $$y = \theta x_1 + (1 - \theta)x_2$$ where $\theta \in \mathbb{R}$, form the line passing through $x_1$ and $x_2$.

I understand intuitively how this is a line where $n = 2$ (i.e. you can write it in slope-intercept form with $m = x_1 - x_2$). But what about for more than two dimensions (i.e. $n \gt 2$)? Why is this guaranteed to be a line geometrically for any $n$?

  • $\begingroup$ Can you picture it in $\mathbb R^3$? $\endgroup$ – J. W. Tanner Feb 25 at 20:50
  • $\begingroup$ I'm not sure I understand what you're asking. Can you clarify? $\endgroup$ – Noah Stebbins Feb 25 at 20:52
  • 2
    $\begingroup$ Does this help? $\endgroup$ – J. W. Tanner Feb 25 at 20:58
  • $\begingroup$ It helps outline the different possible forms. Thank you for the link, @J.W.Tanner. $\endgroup$ – Noah Stebbins Feb 25 at 23:46

The vector $v=x_2-x_1$ is a direction vector of the line.
If, from the origin, we first go to point $x_1$, then take a vector of direction $v$, which is just $\lambda v$ for a scalar $\lambda$, then we arrive to a point on the line: $$x\ =\ x_1+\lambda v\ =\ x_1+\lambda(x_2-x_1)\ =\ (1-\lambda)x_1+\lambda x_2$$ and every point on the line arises this way.
(For swapping $x_1,x_2$, just take $\theta:=1-\lambda$.)


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