Let $V$ be a finite dimensional vector space of dimension $n$. Let $1 \leq k \leq n$ and consider $U_1,...,U_k$ distinct subspaces of $V$, all of dimension $n-1$
a) Prove that $\dim(U_1 \cap U_2 \cap ... \cap U_k) \geq n-k$
b) Give (at least) an example where the equality on $\dim(U_1 \cap U_2 \cap ... \cap U_k) \geq n-k$ doesn't always hold.
My attempt
a) To do the proof, I am using the dimensional theorem
$$\dim(U \cap W) = \dim(U) + \dim(W) - \dim(U + W)$$
And induction.
Let me show you what I have so far.
$\underline{\text{The base case} \ k=1}$
$$\dim(U_1) \geq n-1$$
I think this is not necessary for the proof, but I also checked $k=2$. Using the dimensional theorem I get
$$\dim(U_1 \cap U_2) = \dim(U_1) + \dim(U_2) - \dim(U_1 + U_2)$$
Where
$$\dim(U_1) \geq n-1$$
$$\dim(U_2) \geq n-1$$
$$\dim(U_1 + U_2) \leq n$$
Thus
$$\dim(U_1 \cap U_2) \geq n-2$$
$\underline{\text{Induction step}}$
Here I assumed that $\dim(U_1 \cap U_2 \cap ... \cap U_k) \geq n-k$ holds and proved that it also holds for $k+1$ by means of the dimensional theorem.
$$\dim((U_1 \cap U_2 \cap ... \cap U_k) \cap U_{k+1}) = \dim(U_1 \cap U_2 \cap ... \cap U_k) + \dim(U_{k+1}) - \dim((U_1 \cap U_2 \cap ... \cap U_k) + U_{k+1})$$
Where
$$\dim(U_1 \cap U_2 \cap ... \cap U_k) \geq n-k$$
$$\dim(U_{k+1}) \geq n-1$$
$$\dim((U_1 \cap U_2 \cap ... \cap U_k) + U_{k+1}) \leq n \ \ \text{due to} \ (U_1 \cap U_2 \cap ... \cap U_k) + U_{k+1} \subseteq V$$
Thus
$$\dim((U_1 \cap U_2 \cap ... \cap U_k) \cap U_{k+1}) \geq n-k-1$$
Which is the statement for $k+1$.
$\underline{\text{Conclusion}}$
Since both the base case and the inductive step have been proved as true, by mathematical induction the statement $\dim(U_1 \cap U_2 \cap ... \cap U_k) \geq n-k$ holds for every natural number $k$.
Is my proof OK?
b) Here I did not find any case breaking the equality; I tried with $k=3$, $k=4$ and so on and I always get
$$\dim(U_1 \cap U_2 \cap U_3) \geq n-3$$
$$\dim(U_1 \cap U_2 \cap U_3 \cap U_4) \geq n-4$$
$$...$$
So what am I missing here?
Any help is appreciated.
The proof looks fine.
Consider $\mathbb{R}^3$, and three planes which intersect in a single line. Then $n=3$ and $k=3$, but the dimension of the intersection subspace is $1$.