What is the smallest possible dimension $V_1$ intersection $V_2$ must have.

Question

My answer was $A$ because the two vector spaces may be disjoint and hence intersection is $\{0\}$ whose dimension is zero by convention, but the correct answer is $C$.

Could anyone clarify this for me please?

Answer 1

Not quite, there will necessarily be more intersection due to the size of the vector space the subspaces live in.

If 6 of my dimensions are covered by one of the subspaces, the best I can do (to avoid intersection more than I have to) is fill in the remaining 4 dimensions with a maximally disjoint other subspace, leaving 2 dimensions of necessary intersection.

For an easy to visualize analogue, picture 2 planes in three space. Similar to your problem, you have $$ \dim(V_1)+\dim(V_2)>\dim(V) $$ Can these subspaces intersect only at the origin?

Answer 2

Use the formula $$\dim(V_1\cap V_2)+\dim(V_1+V_2)=\dim(V_1)+\dim(V_2).$$ If $\dim(V_1\cap V_2)=0$ this implies $\dim(V_1+V_2)=12$, which is too big.

Answer 3

Do you know the following formula? \begin{align*} \dim(V_1+V_2)=\dim(V_1)+\dim(V_2)-\dim(V_1\cap V_2) \end{align*} If yes, then \begin{align*} \dim(V_1\cap V_2)&=\dim(V_1)+\dim(V_2)-\dim(V_1+V_2)\\ &=6+6-\dim(V_1+V_2)\\ &\ge 12-\dim(V)\\ &=12-10\\ &=2, \end{align*} where you use that $V_1+V_2$ is a subspace of $V$, hence $\dim(V_1+V_2)\le \dim(V)$.

So the dimension of $V_1\cap V_2$ is always greater than or equal to $2$, which means that the lowest possible dimension is $2$.

Note that this case can indeed occur, namely if $V_1+V_2=V$. (Can you think of an example?)