Why 2 vectors in $\mathbb{R}^{2}$ can be span for the whole dimension, but not in higher dimensions

Question

Lemma: any two vectors in $\mathbb{R}^{2}$ that are not scalar multiples of each other will span all of $\mathbb{R}^{2}$.

why is true for this particular dimension but not necessarily for higher order dimensions?

Answer 1

I am assuming you are referring to using n vectors for the $\mathbb{R}^{n}$ dimension.

The requirement for a set of n vectors to span $\mathbb{R}^{n}$ is that they are linearly independent- that is, no linear combination of all the vectors will equal the zero vector, unless all their coefficients are 0. Thinking graphically, this means that no combination of vectors will end up at the origin (in other words, if you graphically add the vectors, they can't reach where you started.)

For $\mathbb{R}^{2}$, the only way for them to reach the origin (think on a 2-D plane) is if they are scalar multiples of each other. (Try working with it, you won't be able to reach the origin unless they are scalar multiples.) However, for three or more vectors, you can reach the origin by simply making a triangle, or square, or really any set of lines that ends where it starts. The vectors don't have to be scalar multiples of each other, they just have to form a triangle.

As an example, try $\mathbb{R}^{3}$. Given the vectors x = (1,0,0), y = (0,1,0), z = (-1,-1,0), you can't span the dimension using x, y, and z. A quick look shows that any linear combination will yield the third coordinate as 0. Another quick look shows that the vectors are not scalar multiples of each other.

Answer 2

Let $n\geq 3.$ Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ belong to $\Bbb R^n.$

Suppose that for every $z=(z_1,..,z_n)\in \Bbb R^n$ there exist $A,B \in \Bbb R$ such that $$(\bullet)\quad Ax_1+By_1=z_1 \quad \text { and }\quad Ax_2+By_2=z_2.$$ (Note: Some $z\in \Bbb R^n$ would not belong to $\{Ax+By: A,B\in \Bbb R\}$ if this were not so).

Now if $(\bullet)$ is solvable in $A,B$ for every $(z_1,z_2)$ then for every $(z_1,z_2)\ne (0,0)$ there is a $ unique $ pair $ A,B$ that solves $(\bullet).$

Consider $z=(z_1,...z_n)$ and $z'=(z'_1,...,z'_n)$ where $z_1=z_2=z'_1=z'_2=1$ but $z_3\ne z'_3.$ Suppose $Ax+By=z$ and $A'x+B'y=z'.$ Then $A, B$ solves $(\bullet)$ but $A', B'$ also solves $(\bullet).$ So by the uniqueness we have $A=A'$ and $B=B' .$ But then $z=Ax+By=A'x+B'y=z',$ which is absurd , because $z\ne z'.$

So the presumption that $z$ and $z'$ both belong to the span of $\{x,y\}$ is untenable.