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.

Thanks in advance!

  • $\begingroup$ 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 $\endgroup$ – Chinnapparaj R Oct 9 '18 at 7:56
  • $\begingroup$ @ChinnapparajR What if $V=W$? The proposition doesn’t say that the two spaces are distinct. $\endgroup$ – amd Oct 9 '18 at 7:57
  • $\begingroup$ @amd If $V=W$ we surely have inifinitely many vectors in $V \cap W$. $\endgroup$ – Kabo Murphy Oct 9 '18 at 8:00
  • $\begingroup$ Have you learned the identity $\dim(V+W)+\dim(V\cap W)=\dim(V)+\dim(W)$? $\endgroup$ – amd Oct 9 '18 at 8:00
  • $\begingroup$ @ChinnapparajR Indeed, but their intersection isn’t a line as you stated. $\endgroup$ – 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.


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