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Let $U$ and $W$ be subspaces of $\mathbb R^4$ and $u_1=(2,1,3,2), u_2=(-3,0,1,1) \in U$, $w_1=(1,0,1,0),w_2(0,1,0,1) \in W$ and $\dim(U \cap W ^\perp) \ge 2$. I need to find bases for $W, W^\perp, U$.

The vectors $u_1, u_2, w_1, w_2 $ are linearly independent so

  1. $\dim(U) \ge 2 $,
  2. $\dim(W) \ge 2 $,
  3. $\dim(U \cap W^\perp) \ge 2$.

But because of (1) & (3), $\dim(W^\perp) \ge 2 \Rightarrow \dim(W) = 2 $ (because W can not be of one dimension and I know that $W \oplus W^\perp = \mathbb R^4$.

So $W^\perp$ basis will be just $(-1,0,1,0),(0,-1,0,1) $ and so $W $ basis will be just $(1,0,1,0),(0,1,0,1) $

Not sure if I'm right so far.. but if I am.. how do I find a basis for $U$? How can I know if its dimension is $2$ or $3$ or maybe $4$?

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Since $\dim(W^\perp)=2$ and $\dim(U\cap W^\perp)\ge 2$, you have $W^\perp\subseteq U$. Check that the vectors that span $W^\perp$ together with $u_1,u_2$ form a linear independent collection. Conclude that $U=\mathbb R^4$.

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