$\dim(U_1 \cap U_2 \cap U_3) = n − 3$, Give a proof or find a counterexample.

Question

Suppose that $U_1, U_2, U_3$ are three distinct subspaces of $\dim = n-1$ from a vector space of $\dim = n$. where $n \gt 3$.

Give a proof or find a counterexample for $\dim(U_1 \cap U_2 \cap U_3) = n − 3$.

My attempt: first I showed that $\dim(U_1 \cap U_2 \cap U_3) \geq n − 3$, but I do not know what should I do with the inverse. However for $n=3$ I found a counterexample which I mentioned it below.

$U_1 = \{(x,x,y):x,y\in \mathbb{R}\}$ , $U_2 = \{(x,y,x):x,y\in \mathbb{R}\}$ , $U_3 = \{(y,x,x):x,y\in \mathbb{R}\}$ , $U_1 \cap U_2 \cap U_3 = \{(x,x,x):x\in \mathbb{R}\}$

Thank you for your time.

Answer 1

Your counterexample extends to $n = 4$ via $U'_i = \iota(U_i) + \langle e_4 \rangle$ for all $i$ where $\iota\colon \mathbb{R}^3 \hookrightarrow \mathbb{R}^4$ is the inclusion $\iota(x, y, z) = (x, y, z, 0)$. The intersection then satisfies $U'_1 \cap U'_2 \cap U'_3 = \langle e_1 + e_2 + e_3, e_4 \rangle$ and is therefore 2-dimensional.

Answer 2

To construct such $(U_i)_1^3$, we can find a $(n-2)$-dimensional subspace $U$ first and then pick three pairwise independent vectors $v_i$ in the complementary space of $U$ which is feasible since the dimension of the complement is $2$. Then $(span(U,v_i))_1^3$ is a counterexample you want.