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\title[Problem Sets 13 Solutions]{Problem Sets 13 Solutions.}

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\centerline{\sc Problem Set 13: Due 22 September 2000.}

\noindent {\bf Reading.} {\em Matrices and Transformations}, none.

\noindent {\bf Supplementary reading.} Strang, sections 3.1--3.2.

\begin{enumerate}
{\bf \item Consider the set $M_2=\{ 2\times 2\mbox{ matrices}\}$ as a
vector space.  Let
$$
\begin{array}{cc}
A=\left[\begin{array}{rr}2&0\\0&0\end{array}\right] &
B=\left[\begin{array}{rr}0&0\\0&-3\end{array}\right]
\end{array} 
$$
}
\begin{enumerate}
{\bf \item Name a subspace containing $A$ but not $B$.}

All mulitples of the array $A$ form a subspace containing $A$ but not $B$.

{\bf \item Name a subspace containing $B$ but not $A$.}

All multiples of the array $B$ form a subspace containing $B$ but not $A$.

{\bf \item Is there a subspace containing $A$ and $B$ but not the $2\times
2$ identity matrix?}

No.  A subspace must be closed under linear combination.  If a
subspace contains $A$ and $B$, it contains $\frac{1}{2}A -
\frac{1}{3}B$, which is the $2\times2$ identity matrix.

\end{enumerate}

\vspace{.1in}

{\bf \item Consider $\R^2$ as a vector space.  Which of the following are
subspaces and which are not?  If not, why not?}

\begin{enumerate}
{\bf \item $\{ (a,a^2)\ |\ a\in\R\}$}  Not a subspace --- consider $(1,1) + (2,4) = (3,5)$; this set is not closed under addition.
{\bf \item $\{ (b,0)\ |\ b\in\R\}$}  This is a subspace.
{\bf \item $\{ (0,c)\ |\ c\in\R\}$}  This is a subspace.
{\bf \item $\{ (m,n)\ |\ m,n\in\Z\}$}  Not a subspace --- not closed under scalar multiplication: consider $\frac{1}{2} \cdot (1,1)$.
{\bf \item $\{ (d,e)\ |\ d,e\in\R,\ d\cdot e=0\}$ }  Not a subspace --- not closed under addition: consider $(1,0) + (0,1)$.
{\bf \item $\{ (f,f)\ |\ f\in\R\}$ }  This is a subspace.
\end{enumerate}

\vspace{.1in}

{\bf \item Show that for some $b\neq 0$, the solution set $\{ x\ |\ Ax=b\}$ 
does {\bf not} form a subspace.  (Hint: look at Problem set $11$,
problem number $7$.)}

Suppose that the solution set to $Ax=b$ {\em always} forms a subset.
Then, for any two solutions $x_1$ and $x_2$, we have $A(x_1+x_2)=b$,
because subspaces are closed under addition.  But then we have $Ax_1 +
Ax_2 = b$, which, combined with either $Ax_1=b$ or $Ax_2=b$, can be
used to conclude that $Ax_1 = 0$ or $Ax_2 = 0$, an absurdity.  We
conclude that the solution set to $Ax=b$ is not, in general, a
subspace.

\vspace{.1in}

{\bf \item Considerthe set $M_n=\{ n\times n\mbox{ matrices}\}$ as a
vector space.  Which of the following are subspaces?}

\begin{enumerate}
{\bf \item The symmetric matrices, $S=\{ A\ |\ A^{\mathrm{T}}=A\}$}  The symmetric matrices are a subspace.
{\bf \item The non-symmetric matrices, $NS=\{ A\ |\ A^{\mathrm{T}}\neq A\}$}  The non-symmetric matrices are not a subspace --- it is easy to come up with two non-symmetric matrices whose sum is symmetric.
{\bf \item The {\em skew-symmetric} matrices, $S=\{ A\ |\ A^{\mathrm{T}}=-A\}$}  The skew-symmetric matrices are a subspace.
\end{enumerate}

\vspace{.1in}

{\bf \item Describe the column spaces of the following matrices.
$$
\begin{array}{cc}
C=\left[\begin{array}{rr}1&2\\2&0\\-1&3\end{array}\right] &
D=\left[\begin{array}{rrr}1&3&2\\2&2&0\\-1&2&3\end{array}\right]
\end{array}
$$}

\begin{eqnarray*}
C(C) & = & \{c_1(1,2,-1)' + c_2(2,0,3)': (c1,c2) \in \R^2\} \\
C(D) & = & \{c_1(1,2,-1)' + c_2(2,0,3)': (c1,c2) \in \R^2\} (= C(C)) \\
\end{eqnarray*}

Although $D$ has three columns, the column that was added to $C$ to
make $D$ is a linear combination of the two coluns of $C$; the column
spaces of the two matrices are identical.

\vspace{.1in}

{\bf \item Describe the null-space for the following matrices.
$$
\begin{array}{ccc}
E=\left[\begin{array}{rrr}1&2&-2\\2&1&0\end{array}\right] &
F=\left[\begin{array}{rr}1&-1\\0&2\end{array}\right] &
G=\left[\begin{array}{rrr}1&2&-4\\-1&1&3\\1&5&-5\end{array}\right] 
\end{array}
$$
}

\begin{eqnarray*}
N(E) & = & \{c \cdot (2,-4,-3)': c \in \R\} \\
N(F) & = & \{0\} \\ 
N(G) & = & \{c \cdot (10, 1, 3)': c \in \R\} \\
\end{eqnarray*}

\vspace{.1in}

{\bf \item Let $P$ be the plane in $\R^3$ defined by the equation
$$
x-y-z=3.
$$
Find two vectors in $P$ and show that their sum is not in $P$.}

The vectors $(1,0,-2)$ and $(0,1,-4)$ are both in $P$, but their sum,
$(1,1,-6)$, is not in $P$.

\vspace{.1in}

\begin{enumerate} {\item Find a subset $W\subseteq \R^2$ where, for
$v,w\in W$, $v+w\in W$, but $cv$ is not necessarily in $W$.}

We take as our subset all points in the nonnegative orthant, the set
$\{(x,y): x \geq 0, y \geq 0\}$.  The sum of two points in $W$ is
again in $W$, but given a point in $W$, we cannot multiply it by a
{\em negative} scalar to get another point in $W$.

{\bf \item Find a subset $W\subseteq \R^2$ where, for
$v,w\in W$, $cv\in W$, but $v+w$ is not necessarily in $W$.}

We take the union of the two lines $\{(x,y): x = y\}$ and $\{(x,y): x
= -y\}$.  This set is closed under scalar multiplication --- if we
take a point on one of the lines and multiply it by a scalar, we get
another point on the {\em same} line.  However, if we add two points,
each from one of the lines, say $(1,1)$ and $(1,-1)$, we get $(1,0)$,
which is on neither line.

\end{enumerate}

{\bf \item Let $A$ and $B$ be any $n\times n$ matrices.  If $v\in
N(B)$, show that $v\in N(A\cdot B)$.  If $A$ is invertible, show that
if $v\in N(A\cdot B)$, then $v\in N(B)$.}

$$v\in N(B) \rightarrow Bv=0 \rightarrow (A \cdot B)v = A(Bv) = 0
\rightarrow v \in N(A\cdot B)$$

If $A$ is invertible, then $A^{-1}$ exists, and we can write:

\begin{eqnarray*}
v \in N(A \cdot B) & \rightarrow & (A \cdot B)v = 0 \\
& \rightarrow &  A^{-1}(A \cdot B)v = 0 \\
& \rightarrow &  (A^{-1}A)Bv = 0 \\
&  \rightarrow & IBv = 0 \\
& \rightarrow & Bv = 0 \\
& \rightarrow & v \in N(B)
\end{eqnarray*}

\end{enumerate}

\end{document}