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

Question

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 $\dim W\cap V_{t}$

Please verify answer below

Answer 1

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 $