Calculate $\dim W+V$ and $W\cap V$

This is a task from an old exam.

Let define:

$V_{t} = \text{lin}((1,2,2,1),(1,1,-1,t))$

$W = \cases{x_1-x_2=0\\x_3-x_4 =0}$

Calculate $\dim W+V_{t}$ and $\dim W\cap V_{t}$

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Which do you want to know: $W \cap V$ or $W \cup V$? Your questions title and the question itself disagree. –  Fly by Night Jan 21 '13 at 20:56
@FlybyNight Judging by his answer he wants $W \cap V$. –  Git Gud Jan 21 '13 at 20:58
The first one is correct, fixed. Thanks –  Steve Jan 21 '13 at 20:58
Why post a question and then an answer one minute later?! The reply obviously took longer than one minute to type. The intention was clearly to post a question and an answer... I don't get it. –  Fly by Night Jan 21 '13 at 21:00
@FlybyNight Yes, Git Gud is right –  Steve Jan 21 '13 at 21:01

basis of $W$ = $\text{lin} ((1,1,0,0),(0,0,1,1))$

$W+V_{t}=\left[\begin{array}{cccc} 1 & 2 & 2 & 1\\ 1 & 1 & -1 & t\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{array}\right]\sim\left[\begin{array}{cccc} 1 & 2 & 2 & 1\\ 0 & -1 & -3 & t-1\\ 0 & -1 & -2 & -1\\ 0 & 0 & 1 & 1 \end{array}\right]\sim\left[\begin{array}{cccc} 1 & 2 & 2 & 1\\ 0 & 0 & -1 & t\\ 0 & -1 & -2 & -1\\ 0 & 0 & 1 & 1 \end{array}\right]\sim\left[\begin{array}{cccc} 1 & 2 & 2 & 1\\ 0 & -1 & -2 & -1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & t+1 \end{array}\right]$

$\dim W+V_{t}=3$ for $t=-1$ and $4$ in other case.

$V_{t}:$

$\left[\begin{array}{ccc} 1 & 1 & x_{1}\\ 2 & 1 & x_{2}\\ 2 & -1 & x_{3}\\ 1 & t & x_{4} \end{array}\right]=\left[\begin{array}{ccc} 1 & 1 & x_{1}\\ 0 & 0 & x_{2}-2x_{1}\\ 0 & -2 & x_{3}-2x_{1}\\ 0 & t & x_{4}-2x_{1} \end{array}\right]\implies V_{1}:\begin{cases} x_{2}-2x_{1}=0\\ x_{4}-2x_{1}=0 \end{cases}$

$W\cap V_{t}:\begin{cases} x_{1}-x_{2}=0\\ x_{3}-x_{4}=0\\ x_{2}-2x_{1}=0\\ x_{4}-2x_{1}=0 \end{cases}$

$\left[\begin{array}{cccc} 1 & -1 & 0 & 0\\ & & 1 & -1\\ -2 & 1\\ -2 & & & 1 \end{array}\right]\sim\left[\begin{array}{cccc} 1 & -1 & 0 & 0\\ & & 1 & -1\\ 0 & -1\\ 0 & -2 & & 1 \end{array}\right]\sim\left[\begin{array}{cccc} 1 & -1 & 0 & 0\\ 0 & -1\\ & & 1 & -1\\ 0 & -2 & & 1 \end{array}\right]\sim\left[\begin{array}{cccc} 1 & -1 & 0 & 0\\ 0 & -1\\ & & 1 & -1\\ 0 & 0 & & 1 \end{array}\right]$

$\dim W\cap V_{t}=0$

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You seem to apply an equivalence relation on the space of matrices. What does it mean to say that $M \sim N$? Is it actually an equivalence relation? –  Fly by Night Jan 21 '13 at 20:58
I'm following an algorithm to create row reduced (echelon) form of matrix in this steps –  Steve Jan 21 '13 at 21:00
@FlybyNight Using $\sim$ in this context means the following matrix is row or column equivalent to the previous one, which is a way to tidily write down the solution to a system of equations. I didn't read it over but it doesn't look wrong. –  Pedro Tamaroff Jan 21 '13 at 21:03
@Steve "$\dim W+V_{t}=3$ for $t=1$". Did you mean $t=-1$? –  Git Gud Jan 21 '13 at 21:08
@GitGud Yes. Fixed. Thanks –  Steve Jan 21 '13 at 21:10