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\begin{document}
\begin{center}
{\bf Homework 7}\\
(due Wednesday, October 26)
\end{center}
\vspace*{.3in}

\noindent{\bf Book Problems:} 
\begin{description}
\item[3.3, pages 85-86] 3.1, 3.2, 3.3, 3.5 3.10, 3.11 (please convince yourself that the formula in the hint works, but you don't have to actually write down a proof of it)
\item[3.4, page 89] 4.1
\end{description}


\begin{prob} The statement of this problem is a little long, but take heart: it's more windup than pitch.  The 
\emph{product} of two sets $V,W$ is the set
$$
V\times W \eqdef \{(x,y):x\in V, y\in W\}
$$
of all ordered pairs consisting of an element of $V$ followed by one of $W$.
So if $V$ is the set of all planets in the solar system, and $W$ is the set of 
all prime numbers, then $(\text{mars},7)$ would be an example of something in 
$V\times W$.  Now, if by happy coincidence, $V$ and $W$ happen to be vector
spaces over the same field $\F$ of scalars, then $V\times W$ is also a vector 
space over $\F$ in a reasonably natural way.  Given 
$(\vx,\vy),(\vx',\vy')\in V\times W$, one adds according to 
$$
(\vx,\vy)+(\vx',\vy') = (\vx+\vx',\vy+\vy'),
$$
and one multiplies by a scalar $c\in\F$ according to 
$$
c(\vx,\vy) = (c\vx,c\vy).
$$
Verifying that addition and multiplication on $V\times W$ satisfy the vector space axioms is neither difficult nor particularly fun.
Regardless, the axioms all hold, and $V\times W$ actually is a vector space over $\F$.

So for instance, $V$ might be the set of polynomials of degree no larger than 
$3$ and $W$ might be the set of all vectors in $\R^3$ that are perpendicular to
$(1,1,0)$.  Then $V\times W$ is a vector space in which 
$$
(x^2, (1,-1,2)) + (x^3 - x,(0,0,1)) = (x^3 + x^2 - x,(1,-1,3)).
$$
A little unnatural looking perhaps, but it works.
\begin{enumerate}
\item Suppose that $V$ and $W$ are finite dimensional vector spaces.  Show that $\dim V\times W = \dim V + \dim W$ by writing down a basis for $V\times W$ in terms of a basis for $V$ and a basis for $W$.  You needn't prove that your basis actually is a basis as long as it works.  (\emph{Hint: if you need a particular example to think about, consider how the `standard' basis for $\R\times \R$, a.k.a. $\R^2$, is related to the `standard' basis for $\R$.})
\item Suppose now that $W_1,W_2\subset V$ are subspaces of a larger vector
  space $V$.  Verify that the function $T:W_1\times W_2 \to W_1 + W_2$ defined
  by $T(\vw_1,\vw_2) = \vw_1+\vw_2$ is a surjective linear transformation. 
\item Find the kernel of $T$ and describe an isomorphism $S:W_1\cap W_2 \to \ker T$.  You don't need to prove that your isomorphism actually is an isomorphism.
\item Apply the rank theorem to $T$ and use the items above to deduce (as you 
  did by other means a couple of homeworks ago) that
$$
\dim W_1 + W_2 = \dim W_1 + \dim W_2 - \dim W_1\cap W_2.
$$
This strategy for computing the dimension of a vector space is used pretty
often in certain areas of math.
\end{enumerate}
\end{prob}

\begin{prob}
Use the formula for the area of a circle and the relationship between determinants and change of volume to derive the formula for the area inside the ellipse $E := \{(x,y)\in\R^2:\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
\end{prob}




\begin{prob}
Let $A$ be a square matrix with entries in $\R$ and $A^T$ be the transpose of 
$A$.  Then $A$ is called \emph{orthogonal} if $A^TA = I$.  Moreover, $A$ is
called \emph{skew symmetric} if $A^T = -A$
\begin{enumerate}
\item Where do you suppose the term \emph{orthogonal} comes from?
\item Prove that if $A$ is orthogonal, then $\det A = \pm 1$.
\item Prove that if $A$ has an odd number of rows and columns and $A$ is skew
  symmetric, then $\det A = 0$.
\item Are the converses of the previous two statements true?  Prove or give 
  counterexamples.
\end{enumerate}
\end{prob}

\begin{prob}
Let $A$ be a matrix with entries in $\C$.  Let $\bar A$ be the matrix whose
entries are the complex conjugates of those in $A$.  Let 
$A^* = \overline{A^T}$ denote the conjugate transpose of $A$. Then $A$ is
called \emph{unitary} if $A^*A = I$. 
\begin{enumerate}
\item Prove that $\det \bar A = \overline{\det A}$.
\item Prove that if $A$ is unitary, then $|\det A| = 1$.  Note that $\det A$
is complex so $|\det A|$ is the modulus of a complex number rather than the
absolute value of a real number.
\end{enumerate}
\end{prob}

\begin{prob}
If $z_1,\dots,z_n\in \C$, then 
$$
A = \left[\begin{matrix}
1 & z_1 & z_1^2 & \dots & z_1^{n-1} \\
1 & z_2 & z_2^2 & \dots & z_2^{n-1} \\
&     &       & \vdots &   \\
1 & z_n & z_n^2 & \dots & z_n^{n-1} \\
\end{matrix}\right]
$$
is called a \emph{Vandermonde matrix}.  
\begin{enumerate}
\item Verify that if $p(z) = c_0 + c_1 z + \dots + c_{n-1}z^{n-1}$ is a complex
  polynomial, then
$$
A\cdot\cvec{c_0 \\ c_1 \\ \dots \\ c_{n-1}} = 
      \cvec{p(z_1) \\ p(z_2) \\ \dots \\ p(z_n)}.
$$
\item Use this fact to prove that $\det A \neq 0$ if and only if 
$z_1,\dots,z_n$ are distinct.
\end{enumerate}
\end{prob}

\begin{prob}
{\bf BONUS QUESTION (work alone on this one and give solutions direcly to me)}.  Recall that $\Z_2 = \{0,1\}$ is the field 
with only two elements.  Observe that for any given $n$ there are only
finitely many $n\times n$ matrices $A$ with entries in $\Z_2$ (since each such
entry has only two possible values).  Moreover, the determinant of such a
matrix will itself be in $\Z_2$.  So sometimes the determinant will be $0$ and
sometimes it will be $1$.  What fraction of the time will the determinant be $1$?
\end{prob}

\end{document}
\end

