1. ## Subspace combination question

We've done questions a bit like this before, but nothing with subspaces. Basically, for a vector space V there exist subspaces X, Y and Z in V (with dim V > 1), determine, either by proof or counter-example, whether the following statements are true or false:

$\displaystyle X + \left( {Y \cap Z} \right) = \left( {X + Y} \right) \cap \left( {X + Z} \right)$

and

$\displaystyle X \cap \left( {Y + \left( {X \cap Z} \right)} \right) = \left( {X \cap Y} \right) + \left( {X \cap Z} \right)$

Help?!

2. Originally Posted by StandardToaster

We've done questions a bit like this before, but nothing with subspaces. Basically, for a vector space V there exist subspaces X, Y and Z in V (with dim V > 1), determine, either by proof or counter-example, whether the following statements are true or false:

$\displaystyle X + \left( {Y \cap Z} \right) = \left( {X + Y} \right) \cap \left( {X + Z} \right)$
false! counter-example: $\displaystyle V=\mathbb{R}^2, \ X=\{(x,2x): \ x \in \mathbb{R} \}, \ Y=\{(x,0): \ x \in \mathbb{R} \} , \ Z=\{(0,x): \ x \in \mathbb{R}\}.$ obviously $\displaystyle X + Y \cap Z = X,$ because $\displaystyle Y \cap Z=(0).$

now see that $\displaystyle (1,0) \in (X+Y) \cap (X+Z)$ but $\displaystyle (1,0) \notin X=X+Y \cap Z.$ hence the equality does not hold.

$\displaystyle X \cap \left( {Y + \left( {X \cap Z} \right)} \right) = \left( {X \cap Y} \right) + \left( {X \cap Z} \right)$
true by modular law.

3. How does the second one work? I'm utterly confused. I thought I had a proof all written out, but I made a stupid mistake at the start which renders it essentially useless...