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\begin{document}
\section*{Lior Silberman's Math 223: Problem Set 9 (due 22/3/2021)}
Hint for 1,2,3: if you aren't sure try what happens with small matrices
($2\times2,3\times3,4\times4,5\times5$) before tackling the general
case.
\begin{center}
\textbf{Three determinants}
\par\end{center}
\begin{lyxlist}{10.}
\item [{1.}] Fix numbers $a,b$ and let $H_{n}$ be the matrix with entries
$t_{ij}$ so that for all $i$, $t_{ii}=a$, $t_{i,(i-1)}=t_{i,(i+1)}=b$
and $t_{ij}=0$ otherwise. Let $h_{n}=\det H_{n}$.
\begin{lyxlist}{10.}
\item [{(a)}] For $n\geq1$ show that $h_{n+2}=ah_{n+1}-b^{2}h_{n}$.
\item [{(b)}] Using the method of problem 5 below solve the recursion in
the case $a=5$, $b=2$ and find a closed-form expression for $h_{n}$.\bigskip{}
\end{lyxlist}
\item [{2.}] Let $H_{n}(d_{1},\cdots,d_{n})$ be the matrix $J_{n}+\diag(d_{1},\ldots,d_{n})$
where $J_{n}$ is the all-ones matrix and let $h_{n}(d_{1},\cdots,d_{n})=\det\left[H_{n}(d_{1},\cdots,d_{n})\right]$.
\begin{lyxlist}{10.}
\item [{(a)}] Show that $h_{n}(0,d_{2},\ldots,d_{n})=\prod_{j=2}^{n}d_{j}$.
(Hint: subtract the second row from the first)
\item [{(b)}] Suppose that $n\geq2$. Show that $h_{n}(d_{1},d_{2},\ldots,d_{n})=d_{1}h_{n-1}(d_{2},\ldots,d_{n})+d_{2}h_{n-1}(0,d_{3},\ldots,d_{n})$.
\item [{(c)}] Suppose that all the $d_{i}\neq0$ and that $n\geq2$. Show
that $\frac{h_{n}(d_{1},\ldots,d_{n})}{\prod_{j=1}^{n}d_{j}}=\frac{h_{n-1}(d_{2},\ldots,d_{n})}{\prod_{j=2}^{n}d_{j}}+\frac{1}{d_{1}}$.
\item [{(d)}] Show that $\frac{h_{2}(d_{1},d_{2})}{d_{1}d_{2}}=\frac{1}{d_{1}}+\frac{1}{d_{2}}+1$,
and thus that $\frac{h_{n}(d_{1},\ldots,d_{n})}{\prod_{j=1}^{n}d_{j}}=\sum_{j=1}^{n}\frac{1}{d_{j}}+1$.
\item [{CONCLUSION}] $h_{n}(d_{1},\ldots,d_{n})=\left(\sum_{j=1}^{n}\frac{1}{d_{j}}+1\right)\left(\prod_{j=1}^{n}d_{j}\right)$.\bigskip{}
\end{lyxlist}
\item [{3.}] (The ``Vandermonde determinant'') Let $x_{i}$ be variables
and let $V_{n}(x_{1},\ldots,x_{n})$ be the $n\times n$ matrix with
entries $v_{ij}=x_{i}^{j-1}$. We show that $\det V_{n}=\prod_{i=2}^{n}\prod_{j=1}^{i-1}\left(x_{i}-x_{j}\right)$.
\begin{lyxlist}{10.}
\item [{(a)}] Show that $\det V_{n}$ is a polynomial in $x_{1},\ldots,x_{n}$
of total degree $0+1+2+3+\cdots+(n-1)=\frac{n(n-1)}{2}$.
\item [{(b)}] Show that $\det V_{n}$ vanishes whenver $x_{i}=x_{j}$ (which
leads you to suspect that $x_{i}-x_{j}$ divides the polynomial).
\item [{RMK}] Note that $\prod_{i=2}^{n}\prod_{j=1}^{i-1}\left(x_{i}-x_{j}\right)$
is a polynomial of total degree $\frac{n(n-1)}{2}$. It follows from
(a) and the theory of polynomial rings over integral domains that
$\prod_{i=2}^{n}\prod_{j=1}^{i-1}\left(x_{i}-x_{j}\right)$ actually
does divide the determinant, and comparing degrees of the two it follows
that the quotient has degree zero, that is that for some constant
$c_{n}\in\Z$, $\det V_{n}=c_{n}\prod_{i=2}^{n}\prod_{j=1}^{i}\left(x_{i}-x_{j}\right)$.
\item [{SUPP}] Examining the coefficient of $x_{1}^{0}x_{2}^{1}x_{3}^{2}\cdots x_{n}^{n-1}$
show that $c_{n}=1$.
\item [{(d)}] Let $V_{n+1}(x_{1},\ldots,x_{n+1})$ be the matrix described
above, and let $W_{n+1}$ be the matrix obtained by
\begin{lyxlist}{10.}
\item [{(i)}] Subtracting the first row from each row; and then
\item [{(ii)}] For $j$ descending from $n+1$ to $2$, subtracting from
the $j$th column a multiple of the $(j-1)$st so as to make the top
entry in the column zero.
\end{lyxlist}
\item [{~}] Let $\left(w_{ij}\right)_{i,j=1}^{n+1}$ be the entries of
$W_{n+1}$. Show that $w_{11}=1$ that $w_{1j}=w_{i1}=0$ if $i,j\neq1$
and that $w_{ij}=(x_{i}-x_{1})v_{i,j-1}$ if $i,j\geq2$.
\item [{(e)}] Show that $\det V_{n+1}=\left[\prod_{i=2}^{n+1}(x_{i}-x_{1})\right]\cdot\left[\det V_{n}(x_{2},\ldots,x_{n+1})\right]$.
\item [{(f)}] Check that $\det V_{1}=1$ and prove the main claim by induction.\bigskip{}
\end{lyxlist}
\item [{SUPP}] (Polynomial interpolation) Let $\left\{ (x_{i},y_{i})\right\} _{i=1}^{k}\subset\R^{2}$
be points in the plane with distinct $x_{i}$. Show that there exists
a unique polynomial $p\in\R[x]^{0$.
\item [{(e)}] Prove the Theorem.\bigskip{}
\end{lyxlist}
\item [{8.}] Suppose that we add the axiom ``every two distinct lines
intersect at exactly one point''.
\begin{lyxlist}{10.}
\item [{(a)}] Show that in this case exchanging the role of points and
lines (and the adjusting the relation appropriately) gives a new incidence
structure (the ``dual one'') which also satisfies both the axiom
of problem 7 and the axiom we just introduced.
\item [{(b)}] Conclude that with the extra axiom there only three possibilities:
(1) there is exactly one line and it contains all the points; (2)
there is exactly one point and it lies on all lines; (3) there are
as many lines as points\bigskip{}
\end{lyxlist}
\end{lyxlist}
\begin{center}
\textbf{Supplementary problem: Quadratic extensions in general}
\par\end{center}
\begin{lyxlist}{10.}
\item [{A}] (Constructing quadratic fields) Let $F$ be a field, $d\in F$
such that $x^{2}=d$ has no solutions in $F$.
\begin{lyxlist}{10.}
\item [{(a)}] Show that the set of matrices $E=\left\{ \begin{pmatrix}a & b\\
db & a
\end{pmatrix}\mid a,b\in F\right\} $ is a two-dimensional $F$-subspace of $M_{2}(F)$ with basis $1,\epsilon$,
where $\epsilon=\begin{pmatrix} & 1\\
d
\end{pmatrix}$ satisfies $\epsilon^{2}=d$.
\item [{(b)}] Show that $E$ is also closed under matrix multiplication
and transpose.
\item [{(c)}] Show that the map $\sigma\colon E\to E$ given by $\sigma(x)=x^{t}$
satisfies $\sigma(x+y)=\sigma(x)+\sigma(y)$, $\sigma(xy)=\sigma(x)\sigma(y)$,
$\sigma(a+b\epsilon)=a-b\epsilon$ for all $x,y\in E$, $a,b\in F$.
\item [{(d)}] Show that the norm $Nz=z\sigma(z)$ satisfies $Nz\in F$
for all $z\in E$, $Nz\neq0$ if $z\neq0$, $N(zw)=NzNw$.
\item [{(e)}] Conclud that $E$ is a field.
\end{lyxlist}
\item [{B.}] (Uniqueness) Let $E'$ be a field containing $F$ which is
two-dimensional over $F$.
\begin{lyxlist}{10.}
\item [{(a)}] Suppose $E'$ is spanned over $F$ by elements $1,\epsilon$
with $\epsilon^{2}=d$. Let $z=a+b\sqrt{d}\in E'$ be any element
and let $M_{z}\colon E'\to E'$ be the map of multiplication by $z$.
Show that $M_{z}$ is $F$-linear and that its matrix in the basis
$\left\{ 1,\epsilon\right\} $ is $\begin{pmatrix}a & b\\
db & a
\end{pmatrix}$.
\item [{(b)}] Show that $E$ always has a basis of the form $\left\{ 1,\delta\right\} $
with $\delta\notin F$. Show that if $\chr F\neq2$ there is $\epsilon=a+b\delta$
such that $\epsilon^{2}\in F$.
\item [{(c)}] Show that $E=F(\sqrt{d})$ and $E'=F(\sqrt{d'})$ are isomorphic
as fields iff $\frac{d}{d'}$ is a square in $F$.
\end{lyxlist}
\end{lyxlist}
\end{document}