# Intersection of two subspaces of a 3 dimensional vector space

I have this transformation problem which is sort of a homework problem and the thing is that it is quite troublesome. So, basically the question has two parts

1. Prove that if V and W are two 2 dimensional subspaces of $$\mathbb{R}^{3}$$, then there exists infinitely many vectors in $$V\cap W$$.

For this I proceeded like this; Let the basis of V be $$\{v_{1}, v_{2} \}$$ and that of W be $$\{w_{1}, w_{2} \}$$, then their intersection will have the basis $$\{v_{1}, v_{2}, w_{1}, w_{2} \}$$. Since they are the subspaces of $$\mathbb{R}^{3}$$, this combination of bases must span $$\mathbb{R}^{3}$$? Also, if it does so, anyone of the vectors in this basis could be achieved as a linear combination of the other three, because the basis of $$\mathbb{R}^{3}$$ cannot contain four vectors. This is point from where I could not continue any more.

1. Can we define a linear transformation T: $$\mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$$ whose kernel, i.e., $$ker(T)=V$$ and $$image(T)=W$$? I know that the kernel is that set of vectors for which the transformation yields a zero vector. So if I am interpreting the kernel subpart of this question correct, then it is trying to say that this 2 dimensional subspace V gets transformed to a zero vector, that is origin when the transform is applied, i.e., a plane is transformed to a point in $$\mathbb{R}^{3}$$. I have no idea what to do next and for the second subpart (Image part), I have no clue at all. Pardon me if I am asking too much. The least anyone could do is to provide some hints in both the parts 1 and 2, if not the whole solution.

I request for a simpler explanation. I don't have any background of linear algebra before.

• Two dimensional subspaces of $\Bbb{R}^3$ are precisely plane through the origin and so their intersection is a line through the origin if they intersect – Chinnapparaj R Oct 9 '18 at 7:56
• @ChinnapparajR What if $V=W$? The proposition doesn’t say that the two spaces are distinct. – amd Oct 9 '18 at 7:57
• @amd If $V=W$ we surely have inifinitely many vectors in $V \cap W$. – Kabo Murphy Oct 9 '18 at 8:00
• Have you learned the identity $\dim(V+W)+\dim(V\cap W)=\dim(V)+\dim(W)$? – amd Oct 9 '18 at 8:00
Here is a simpler method. If $$V \cap W$$ has one non-zero vector it has infinitely many vectors (since scalar multiples of the vector belong to $$V \cap W$$). Suppose it has no non-zero vector. Then the dimension of $$V+W$$ would be $$2+2=4$$ but no subspace of $$\mathbb R^{3}$$ can have dimension $$4$$. That finishes the proof.