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\title[Problem Set 15 Solutions]{Problem Set 15 Solutions}

\begin{document}

\maketitle

\centerline{\tiny by Michael Allen}

\begin{enumerate}

\item Which of the following sets of vectors span $\R^3$?
\begin{enumerate}
\item $(1,2,0$ and $(0,-1,1)$.

    {\bf No.} Two vectors cannot span $\R^3$.

\item $(1,1,0)$, $(0,1,-2)$, and $(1,3,1)$.

    {\bf Yes.} The three vectors are linearly independent, so they
    span $\R^3$.

\item $(-1,2,3)$, $(2,1,-1)$, and $(4,7,3)$.

    {\bf No.} Only two of these vectors are linearly
    independent, and cannot span $\R^3$.

\item $(1,0,2)$, $(0,1,0)$, $(-1,3,0)$, and $(1,-4,1)$.

    {\bf Yes.} Three of these vectors are linearly independent,
    so they span $\R^3$.

\end{enumerate}
\medskip

\item Which of the following sets of vectors span $P_3=\{
at^3+bt^2+ct+d\}$?
\begin{enumerate}
\item $t+1$, $t^2-t$, and $t^3$.

    {\bf No.} The space is 4-D, but there are only three vectors.

\item $t^3+t$ and $t^2+1$.

    {\bf No.} Again, there are not enough vectors to span a 4-D space.

\item $t^2+t+1$, $t+1$, $1$, and $t^3$.

    {\bf Yes.} These vectors are linearly independent and span $P_3$.

\item $t^3+t^2$, $t^2-t$, $2t+4$, and $t^3+2t^2+t+4$.

    {\bf No.} The fourth vector is the sum of the first three, so they cannot span $P_3$.

\end{enumerate}
\medskip

\item Are the following sets of vectors linearly dependent or
independent?  If they are dependent, write one as a linear combination of
the others.

\begin{enumerate}
\item $(1,2,0)$ and $(0,-1,1)$ in $\R^3$.

    {\bf Independent.}

\item $(-1,2,3)$, $(2,1,-1)$, and $(4,7,3)$ in $\R^3$.

    {\bf Dependent.} $2(-1,2,3) + 3(2,1,-1) = (4 7 3)$

\item $(1,2)$, $(2,3)$, and $(8,-2)$ in $\R^2$.

    {\bf Dependent.} $18(2,3) - 28(1,2) = (8,-2)$

\item $t^2+2t+1$, $t^3-t^2$, $t^3+1$, and $t^3+t+1$ in $P_3$.

    {\bf Dependent.} $\frac{1}{2}(t^2+2t+1) + \frac{1}{2}(t^3-t^2) + \frac{1}{2}(t^3+1) = t^3+t+1$
\end{enumerate}
\medskip

\item What is the dimension of the following spaces?
\begin{enumerate}
\item The set of $2\times 2$ symmetric matrices, $A=A^T$.
$$A=\left[\begin{array}{cc}a&b\\b&d\end{array}\right],$$

    {\bf Three.} Since the upper-left and lower-right corners must
    be equal, there are only three variables which can be changed.

\item The set of $2\times 2$ matrices
$$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right],$$
with $a+d=0$.

    {\bf Three.} We can rewrite the constraint as
$$A=\left[\begin{array}{cc}a&b\\c&-a\end{array}\right],$$
    which makes it clear that the space is three dimensional.

\item The set $\{ (x,y,x-3y,2y-x)\ |\ x,y\in\R\}$ inside of $\R^4$.

    {\bf Two.} Since only x and y are free, this space is two
    dimensional.

\end{enumerate}

\medskip

\item What is the column space and row space of the matrix
$$A=\left[\begin{array}{rrrr}1&3&5&-2\\2&-1&3&-4\\-1&4&2&2\end{array}\right]?$$

{\bf Answer:} If we use Gaussian Elimination on the matrix, we
find that there are only two pivots, so we know both the column
and row spaces are two dimensional. So, let's just take the first
two columns and the first two rows (as long as they are linearly
independent of each other).

$$\begin{array}{cc}
    x_1\left[\begin{array}{c}1\\2\\1\end{array}\right]
    + x_2\left[\begin{array}{c}3\\-1\\4\end{array}\right]
    &
    x_1\left[\begin{array}{cccc}1&3&5&-2\end{array}\right]
    + x_2\left[\begin{array}{cccc}2&-1&3&-4\end{array}\right]
\end{array}$$

\medskip

\item Find an (infinite) basis for the space of all polynomials
$$\mathcal{P}=\{ a_nx^n+z_{n-1}x^{n-1}+\cdots+a_1x+a_0\ |\ \mbox{ for all }
n\}.$$

{\bf Answer:} $\{ a_nx^n | n = 0,1,2...\}$

\medskip

\item Suppose $\{ v_1,\dots,v_n\}$ spans a vector space $V$, and suppose
that $v_n$ is a linear combination of $v_1$ through $v_{n-1}$.  Then show
that $\{ v_1,\dots,v_{n-1}\}$ spans $V$ as well.

{\bf Answer:} Since we can add to our span any vector which is a
linear combination of the vectors already in the span, if follows
easily that the two sets of vectors are equivalent, and $\{
v_1,\dots,v_{n-1}\}$ must span $V$.

\medskip

\item If $A$ is a $4\times 6$ matrix, show that the columns of $A$ are
linearly dependent.

{\bf Answer:} Since there are only 4 rows, the matrix can at most
be of rank 4. There are 6 columns in the matrix and only 4 of
them can be linearly independent, so at least two columns must be
linear combinations of the others.

\medskip

\item Compute $$\left[\begin{array}{rr}.1&.95\\.9&.05\end{array}\right]^n,$$
for $n=3$, $5$ and $100$ using methods from recitation.

{\bf Answer:} We begin this problem by finding the eigenvalues and
eigenvectors of the array from which will will construct our
answer. These turn out to be:

$$\begin{array}{cc}
    \lambda_1 = 1 & \lambda_2 = -0.85\\
    v_1 = \left[\begin{array}{c}19\\18\end{array}\right] &
    v_2 = \left[\begin{array}{c}1\\-1\end{array}\right]
  \end{array}$$

We decompose the columns of the matrix into eigenvectors, do a
lot of algebra, and mathemagically, we arrive at the following
answers:

$$\begin{array}{ccc}
    A^3=\left[\begin{array}{cc}0.215&0.829\\0.785&0.171\end{array}\right]&
    A^5=\left[\begin{array}{cc}0.298&0.741\\0.702&0.259\end{array}\right]&
    A^{100}=\left[\begin{array}{cc}0.514&0.514\\0.486&0.486\end{array}\right]
  \end{array}$$
\end{enumerate}

\end{document}