Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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 $

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.