I'm watching the video tutorial on spans here: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/linear_combinations/v/linear-combinations-and-span

At 8:13, he says that the vectors a = [1,2] and b = [0,3] span R2. Visually, I can see it. But I tried to work it out, like so:

sp(a, b) = x[1,2] + y[0,3] such that x,y exist in R
         = [x, 2x] + [0, 3y] st x,y e R
         = [x, 2x + 3y] st x,y e R

With that said, how do we know that [x, 2x + 3y]spans R2? I tried picking a random point ([19, 6]) and let x=19 and solved for y (2*19 + 3y = 6) and found that when x=19 and y=-10, then I can get the point [19, 6].

But I'm confused as to how would I be able to find out that [x, 2x + 3y] can make any and all points in R2? What's my next step in determining if it spans R2?


Well then, let us pick any point in $\mathbb R^2$, e.g. the point $(a,b)$.

Let us define $x = a$ and $y=\frac{b-2a}{3}$ then you can easily see that:

$$ \begin{align*} (a,[b-2a]/3) = (x,y) \\ \iff (a,b-2a) = (x,3y) \\ \iff (a,b) = (x,2a+3y) \\ \iff (a,b) = (x,2x+3y) \end{align*} $$

So $(a,b)$ really is in the span of $\{(1,2),(0,3)\}$.

As we can choose any $(a,b) \in \mathbb R^2$, we know that those vectors span the whole $\mathbb R^2$.

  • $\begingroup$ Thanks for the response. Quick question, how exactly did we get y = b-2a / 3? $\endgroup$ – user2719875 Jan 16 '16 at 23:46
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    $\begingroup$ I started from the last line=) (Many times in math the proof goes in the other direction compared to the order of the discovery.) Assume there is $x,y$ such that $(a,b) = (x,2x+3y)$.. Can we solve for $(x,y)$ or do we get a contradiction? $\endgroup$ – flawr Jan 17 '16 at 9:56

For any $(a,b)\in \mathbb{R}^2$, $$(a,b) = a\cdot(1,2)+(-\frac{2a}{3}+\frac{b}{3})\cdot(0,3)$$

  • $\begingroup$ Thanks for the response. Quick question, for any (a, b) ∈ R2, doesn't (a, b) = a(1, 2) + b(0, 3)? Where did we get (−2a/3 + b/3) from? $\endgroup$ – user2719875 Jan 16 '16 at 23:49
  • $\begingroup$ If we simplify the RHSwe get $$(a,2a) +(0, -2a+b) = (a,b)$$. Does this make more sense? All we did was multiple the scalars into the vectors and then add component-wise. If we used your suggestion we end up with $$(a,b) = (a,2a+3b)$$ Which is only true if $$b=0$$. $\endgroup$ – fosho Jan 16 '16 at 23:52
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    $\begingroup$ Right, but how exactly did you arrive at (a,2a)+(0,−2a+b)=(a,b) (or the equation in your answer) in the first place? $\endgroup$ – user2719875 Jan 17 '16 at 0:07

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