# $Dim(W_{1}\cap W_{2}\cap W_{3})$, for given $9$ dimension sub spaces $W_{1},W_{2}$ and $W_{3}$ of $\mathbb{R}^{10}.$

Let $W_{1},W_{2}$ and $W_{3}$ be three distict subspaces of $\mathbb{R}^{10}$ such that each of $W_{i}$ has dimension $9.$ Let $W=W_{1}\cap W_{2}\cap W_{3}.$ Then we can conclude that

$1.$ $W$ may not a subspace of $\mathbb{R}^{10}.$

$2.$ $dim(W)\leq8.$

$3.$ $dim(W)\geq7.$

$4.$ $dim(W)\leq3.$

It is obvious option first and fourth are not true. But to handle option second and third. Please help.

The only way $\dim W \ge 9$ can happen is if $W_1=W_2=W_3$, but this contradicts distinctness of the $W_i$. So #2 need not be true.
Using the identity $\dim(U+V)=\dim(U)+\dim(V)-\dim(U \cap V)$ for two subspaces (note: this does not generalize à la inclusion-exclusion to more than two subspaces!) we have $$10=\dim(\mathbb{R}^{10})=\dim(W_1+W_2)=9+9-\dim(W_1 \cap W_2)$$ so $\dim(W_1 \cap W_2)=8$, where we use distinctness of $W_1$ and $W_2$ to conclude $W_1+W_2=\mathbb{R}^{10}$.
Applying it again gives $$10 \ge \dim((W_1 \cap W_2) + W_3) = 8 + 9 - \dim(W_1 \cap W_2 \cap W_3)$$ which proves #3 is true.
• my be distinct with dim as $9.$ – neelkanth Dec 20 '16 at 11:03
• it says that dimension $\leq 8$ – user464147 Jun 13 '19 at 7:52