I am trying to prove that a set of two vectors in $\mathbb{R}^3$ cannot be a spanning set for $\mathbb{R}^3$.
Intuitively , it makes a lot of sense, however I need to prove it for my homework assignment. This is for a proof-based linear algebra class.
Edit:
Apologies,
This far I have tried to prove that in general we cannot find a set of constants that satisfies the spanning definition.
Essentially taking 2 vectors $v_1, v_2 \in \mathbb{R}^3$ and saying for these vectors to span $\mathbb{R}^3$, $\exists \alpha_1, \alpha_2$ such that $\alpha_1\cdot v_1 + \alpha_2\cdot v_2=x$ where x is any arbitrary vector in $\mathbb{R}^3$.
I then took the components of the vectors $v_1$ and $v_2$ and set them equal to the components of the arbitrary vector $x$. That will yield a system of equations. However I’m stuck and cannot figure out if this is even the right approach.
I don’t think it is enough for me to simply say "since the vector space has 3 dimensions we need at least 3 vectors with three components to span it" without proof.
Hope that’s better.
Best Answer
Let us try Contradiction.
When we can span $R^3$ with 2 vectors from $R^3$ , we must atleast get $V_1=(1,0,0),V_2=(0,1,0),V_3=(0,0,1)$ which can then easily generate all other vectors.
Hence we have to show that we can not get $V_1,V_2,V_3$ through Contradiction.
Assume we can get those 3 vectors using 2 vectors $U_1=(u_{11},u_{12},u_{13}),U_2=(u_{21},u_{22},u_{23})$.
$r_{11}U_1+r_{12}U_2=V1 \tag{1}$
$r_{21}U_1+r_{22}U_2=V2 \tag{2}$
$r_{31}U_1+r_{32}U_2=V3 \tag{3}$
With (1) , where the last component is $0$ , we get : $r_{11}u_{13}+r_{12}u_{23}=0 \tag{4}$
With (2) , where the last component is still $0$ , we get :
$r_{21}u_{13}+r_{22}u_{23}=0 \tag{5}$
Subtracting the two , after cross multiplying , we can eliminate $u_{13}$ to get $r_{12}r_{21}u_{23}-r_{11}r_{22}u_{23}=0$ & we can eliminate $u_{23}$ to get $r_{11}r_{22}u_{13}-r_{12}r_{21}u_{13}=0$
This indicates that either (A) $u_{13}=u_{23}=0$ or (B) $r_{12}r_{21}-r_{11}r_{22}=0$
We see that Case (A) indicates that third component can never go Non-Zero to generate $V_3$ , hence Contradiction.
We can then check that Case (B) indicates that $V_2$ is always a multiple of $V_1$ , hence Contradiction.
[[
this answer seems too long , though the core idea is very simple
somebody else might give shorter arguments using some theorems
]]