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Consider the subspaces $W_1$ and $W_2$ of $\mathbb{R}^3$ given by $W_1= \{(x,y,z) \in \mathbb{R}^3:x+y+z=0 \}$ and $W_2=\{(x,y,z) \in \mathbb{R}^3:x-y+z=0 \}$.

If $W$ is a subspace of $\mathbb{R}^3$ such that

  • $W \cap W_2= \mathrm{span}\bigl\{(0,1,1)\bigr\}$

  • $W \cap W_1$ is orthogonal to $W \cap W_2$ with respect to the usual inner product of $\mathbb{R}^3$

then which of these are true?

  1. $W = \mathrm{span} \bigl\{ (0,1,-1),(0,1,1) \bigr\}$

  2. $W = \mathrm{span} \bigl\{ (1,0,-1),(0,1,-1) \bigr\}$

  3. $W = \mathrm{span} \bigl\{ (1,0,-1),(0,1,1) \bigr\}$

  4. $W = \mathrm{span} \bigl\{ (1,0,-1),(1,0,1) \bigr\}$

My Attempt:
$x+y+z=0 \implies x+y=-z$ so that free variables are two so $\mathrm{dim}(W_1)=2$ and similarly $x-y+z=0 \implies x+z=y$ so that $\mathrm{dim}(W_2)=2$.

Also $W \cap W_2 = \mathrm{span}\bigl\{(0,1,1) \bigr\}$ implies $(0,1,1)$ is one element of $W$ so options 2,4 discarded.

How to approach this type of problems in general?

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    $\begingroup$ Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Dave
    Commented Dec 24, 2018 at 23:35
  • $\begingroup$ I edited it but I use the symbol $ then curly braces removed. $\endgroup$
    – Mathforjob
    Commented Dec 25, 2018 at 0:40
  • $\begingroup$ You have to "escape" the braces, by typing \{, since they are usually used for something else. $\endgroup$
    – user403337
    Commented Dec 25, 2018 at 1:09
  • $\begingroup$ Thanks for the hint $\endgroup$
    – Mathforjob
    Commented Dec 25, 2018 at 5:57

1 Answer 1

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You reasoned correctly and discarded $2$ and $4$. It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.

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  • $\begingroup$ Thank you for the help $\endgroup$
    – Mathforjob
    Commented Dec 25, 2018 at 6:52

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