The linear relationship between two plane(or two subspace)

Question

Suppose $V$ is a $3$-dimensional vector space, suppose I have $2$ planes $S_1$ and $S_2$ going through the origin, then they must at least intersect at a line.

Suppose $V$ is a $4$-dimensional vector space, suppose I have $2$ planes $S_1$ and $S_2$ going through the origin, then they must at least intersect at the origin.

Then the more general question is, if I have an $n$-dimensional vector space $V$, and I have two subspaces $S_1$ and $S_2$ with dimension $r$, their intersection forms an at least dimension $(n-r)$ subspace.

I'm trying to prove this: first take $S_1$ and $S_2$ as $2$ dimensional subspaces (a plane), then choose linearly independent $v_1, v_2$ to form $S_1$, then choose linearly independent $v_3, v_4$ to form $S_2$, then their intersection is $a_1v_1 + a_2v_2 = a_3v_3 + a_4v_4$.

$a_1v_1 + a_2v_2 - a_3v_3 - a_4v_4 = 0$.

Suppose $V$ is an $n$-dimensional space, then there's $(n-r)$ redundant $v$ here that can be expressed as other $v$'s linear combination, then I'm blocked here. How do I say that their intersection is at least dimension $(n-r)$ subspace? Also I can't find the equation to describe the intersection subspace.

Answer 1

I think what you want is to show that for a finite-dimensional vector space $V$ and subspaces $S_1, S_2$, $\text{dim }(S_1 \cap S_2) \geq \text{dim } S_1 + \text{dim } S_2 - \text{dim } V$. This can be shown using the proof that $\text{dim }(S_1 \cap S_2) = \text{dim } S_1 + \text{dim } S_2 - \text{dim } (S_1 + S_2)$ and the fact that since $S_1,S_2 \subseteq V$, $S_1 \cup S_2 \subseteq V$, so $\text{dim } (S_1 + S_2) \leq \text{dim }V$. This explains why if $\text{dim } V = n$ and $\text{dim } S_1 = \text{dim } S_2 = r$, $\text{dim } (S_1 \cap S_2) \geq 2r - n$. In your specific case, in a three-dimensional vector space ($n=3$), the intersection of two planes ($r=2$) is at least a line ($2r-n=1$). (The fact that this also equals $n-r$ is pure coincidence.)

Regarding finding the equation, this is equivalent to solving a large system of linear equations. See this post for an example. In many ways, solving these linear equations is a huge point of linear algebra (think Ax = b), so you're not always going to have a quick and easy equation.

Answer 2

This is not true. Take any two distinct lines in $\mathbb R^n$ for $n\geq 2$. Their intersection is $\{0\}$, so has dimension $0$, which is less than $n-1$.