308F/G Winter 2013, Quiz 6

Each class got two of the following four problems

Invert the following matrix: $A=\left(\begin{array}{ccc} 1 & 4 & 2 \\ 0 & 2 & 1 \\ 3 & 5 & 3 \end{array}\right)$. To do this, we arrange the matrix along with the 3x3 identity matrix and row reduce: \[(A|I)=\left(\begin{array}{ccc|ccc} 1 & 4 & 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 3 & 5 & 3 & 0 & 0 & 1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 4 & 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 0 & -7 & -3 & -3 & 0 & 1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 4 & 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & -3 & 4 & 1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 0 & -2 & 13 & -16 & -4 \\ 0 & 0 & -1 & 6 & -7 & -2 \\ 0 & 1 & 1 & -3 & 4 & 1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & -6 & 7 & 2 \\ 0 & 1 & 0 & 3 & -3 & -1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -2 & 0 \\ 0 & 1 & 0 & 3 & -3 & -1 \\ 0 & 0 & 1 & -6 & 7 & 2 \\ \end{array}\right) \] Hence, $A^{-1}=\left(\begin{array}{ccc} 1 & -2 & 0 \\ 3 & -3 & -1 \\ -6 & 7 & 2 \\ \end{array}\right)$.
Find the inverse of $A= \left(\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & 4 \\ 0 & 2 & 7 \end{array}\right).$ Again, we row reduce \[(A|I) =\left(\begin{array}{ccc|ccc} 1 & 3 & 5 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 2 & 7 & 0 & 0 & 1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 0 & -7 & 1 & -3 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & -2 & 1 \end{array}\right) =\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 11 & -7 \\ 0 & 1 & 0 & 0 & -7 & 4 \\ 0 & 0 & 1 & 0 & 2 & -1 \end{array}\right)\] and so the inverse is $A^{-1}=\left(\begin{array}{ccc} 1 & 11 & -7 \\ 0 & -7 & 4 \\ 0 & 2 & -1 \end{array}\right)$.
This is problem (a) in the extra problems (note, you can use this fact on your exam and homeworks, I just wanted you guys to know how to prove this so you understood why it was true). Suppose $W\subseteq V$ and $\dim V=\dim W$, show $V=W$. Let $w_{1},...,w_{n}$ be a basis for $W$, then $n=\dim W=\dim V$, and $w_{1},...,w_{n}\in V$. Thus, we have $n$ linearly independent vectors $w_{1},...,w_{n}$ in $V$, where $n=\dim V$, which implies they are a basis for $V$, hence \[W=Sp\{w_{1},...,w_{n}\}=V.\]
This is problem b in the extra problems. If $N(T)=\{0\}$, show that if $T(u)=T(v)$, then $u=v$. Well, if $T(u)=T(v)$, then \[0=T(u)-T(v)=T(u-v)\] and so $u-v\in N(T)=\{0\}$, which implies $u-v=0$, i.e. $u=v$.
This is half of problem c in the extra problems. Show that if $T:V\rightarrow W$ is linear, then $N(T)$ is a subspace of $V$. There are three things we need to verify:
- Observe that $T(0)=0$ because $T$ is linear, and so $0\in N(T)$.
- Let $x,y\in N(T)$, so $T(x)=T(y)=0$. Then $T(x+y)=T(x)+T(y)=0+0=0$, and so $x+y\in N(T)$.
- If $x\in N(T)$ and $c\in \mathbb{R}$, then $T(cx)=cT(x)=c\cdot 0=0$, and so $cx\in N(T)$.
Thus, $N(T)$ is a subspace.
This is problem e in the extra problems. Suppose $T:V\rightarrow W$ is linear and $\dim W+nullity(T)=\dim V$, show $R(T)=W$. By the rank-nullity theorem, \[\dim W=\dim V-nullity(V)=rank(T),\] and so $R(T)$ is a subspace of $W$ with $\dim R(T)=\dim W$, and hence $R(T)=W$ (recall part (a) of the extra problems).