308F/G Winter 2013, Quiz 1

  1. The first problem was problem 35 from section 1.2. You needed to solve the equations \[x_{1}-x_{2}-x_{3}=1\] \[x_{1}+x_{3}=2\] \[x_{2}+2x_{3}=3.\] The first step is to create the augmented matrix for this system, which is \[\left(\begin{array}{ccc|c} 1 & - 1 & -1 & 1 \\ 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 3 \end{array}\right).\] Then we do row operations. We first subtract the first row from the second, obtaining \[\left(\begin{array}{ccc|c} 1 & - 1 & -1 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 2 & 3 \end{array}\right).\] Then we subtract the second row from the third to get \[\left(\begin{array}{ccc|c} 1 & - 1 & -1 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{array}\right).\] At this point, we can stop reducing the matrix and observe that the last row reads off the equation $0=1$, which is impossible, therefore the system of equations is inconsistent and has no solutions.
  2. This is problem 30 from section 3.2 in the book. It asks to show that, if $U$ and $V$ are subspaces, then the set \[U+V=\{x:x=u+v,\; u\in U, \; v\in V\}\] is a subspace. This is the set of all vectors $x$ that can be written as something in $U$ plus something in $V$. To show this is a subspace, we verify the three conditions:
    1. $0\in U+V$. This follows because $0\in U$ and $0\in V$, so that $0=0+0$, that is, $0$ is something in $U$ plus something in $V$, so $0\in U+V$.
    2. $x,y\in U+V$ implies $x+y\in U+V$. To see this, note that since $x\in U$, $x=u+v$ for some $u\in U$, $v\in V$, and $y=s+t$ for some $s\in U$ and $t\in V$, hence \[x+y=(u+v)+(s+t)=\underbrace{u+s}_{\in U} + \underbrace{v+t}_{\in V} \in U+V\] where we used the fact that $u+s\in U$ (since $u,s\in U$) and $v+t\in V$ (since $v,t\in V$).
    3. $cx\in U+V$ for any $x\in U+V$ and $c\in \mathbb{R}$. Since $x\in U+V$, $x=u+v$ for some $u\in U,V\in V$, hence \[cx=c(u+v)=\underbrace{cu}_{\in U}+\underbrace{cv}_{\in V}\in U+V.\] This finishes the proof.