Finding a basis for a subspace in $\mathbb{R}^2$

Question

We are given two vectors $(a,b),(c,d)\in\mathbb{R}^2$ which lie in the same one dimensional subspace, and at least one of $a,b,c,d$ is nonzero. How can we find a single vector in terms of $a,b,c,d$ which spans the subspace containing the two original vectors?

Taking the average, or in fact any linear combination of $(a,b),(c,d)$, seems the obvious answer at first. However, one can always find pairs of vectors in the same one dimensional subspace for which this linear combination would be the zero vector, the span of which is clearly not the wanted subspace. There should be an easy trick but I am not seeing it.

Answer 1

Withour lost of generality, assume that either $a$ or $b$ are non-zero. Since the two vectors lie in the same 1-dimensional vector space, then, there exists $\lambda\in\mathbb R$ such that $(c,d)=\lambda (a,b)$. Now I distinguish three cases:

• If $\lambda=0$, then $c=d=0$ and the sum (which is just the first vector) spans the vector subspaces in terms of $a,b,c$ and $d$.

• If $\lambda>0$, the sum of both vectors spans the vector subspace in terms of $a,b,c$ and $d$.

• Finally, if $\lambda<0$, the difference is considered.

Answer 2

Since the vectors $(a,b)$ and $(c,d)$ lie in the same one dimensional subspace of $\mathbb{R}^2$, then they lie on the same line, which means $(a,b), (c,d)$ are scalar multiples of each other, or $(a,b) = \lambda (c,d)$ for some $\lambda$. This means that a basis for our subspace is just $(a,b)$, since if $\alpha(a,b) + \beta(c,d)$ is a linear combination of the original vectors, we can write this as $\alpha\lambda(c,d) + \beta(c,d) = (\alpha \lambda + \beta)(c,d)$, so any vector in the subspace is in the span of $(c,d)$.