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\begin{document}
\centerline{Math 201 Homework for Friday, Week 8}
\bigskip
Due: Friday, October 29.

\begin{prob}
  Compute the determinant of the following matrices by using row operations. 
  \begin{enumerate}[(a),itemsep=12pt]
      \item $\begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8  \\ 3 & 6 & 9  \end{pmatrix}$
      \item $\begin{pmatrix} 1 & 3 & -1 & 2\\ 2 & 4 & 7 & -3 \\ 0 & 0 & 2 &
       0\\ 0& 0 & 0 & 6 \end{pmatrix}$
     \item $\begin{pmatrix} 4 & -1 & -1 & -1\\ -1 & 4 & -1 & -1 \\ -1 & -1 & 4 &
       -1\\ -1& -1 & -1 & 4 \end{pmatrix}$
       \item (BONUS) Generalize part (c) for the $n\times n$ matrix $\begin{pmatrix} n & -1 & \cdots & -1\\ -1 & n & \cdots & -1 \\ \vdots & \vdots & \ddots &
       \vdots\\ -1& -1 & \cdots & n \end{pmatrix}$, with $n$ in the main diagonal and $-1$ everywhere else.
    \end{enumerate}
  
\end{prob}

\begin{prob}
      Let~$V$ be a vector space.  For each integer~$r>0$, we now give a
      provisional definition of a new vector space called~$\bigwedge^rV$. A
      spanning set for~$\bigwedge^rV$ consists of expressions of the
      form~$v_1\wedge\dots\wedge v_r$ where~$v_1,\dots,v_r\in V$.  For example,
      if~$u,v,w\in V$, the following would be typical elements of~$\bigwedge^2V$:
      \begin{align*}
        \omega &= 4\,u\wedge v-5\,u\wedge w+7\,v\wedge w\\
        \mu &= 2\,u\wedge v+9\,u\wedge w+6\,v\wedge w.
      \end{align*}
      Addition is done by combining like terms, and scaling is done by scaling
      each term.  For instance, continuing the example above, we get
      \begin{align*}
        \omega+\mu&=6\,u\wedge v+4\,u\wedge w+13\,v\wedge w\\
        5\,\omega &=20\,u\wedge v-25\,u\wedge w+35\,v\wedge w.
      \end{align*}
      We now add a couple of rules.  First, these ``wedge products'' of vectors
      are
      linear in each component.  For~$v_i\in V$ and~$a\in F$,
      \begin{align*}
        &v_1\wedge\dots\wedge v_{i-1}\wedge(a v_i+v_i')\wedge v_{i+1}\wedge\dots
        v_r=\\[12pt]
        &\hspace{1in} a\,v_1\wedge\dots\wedge v_{i-1}\wedge v_i\wedge
        v_{i+1}\wedge\dots\wedge v_r\\
        &\hspace{2in}+\\
       &\hspace{1in} v_1\wedge\dots\wedge v_{i-1}\wedge v_i'\wedge
       v_{i+1}\wedge\dots\wedge v_r.
      \end{align*}
      Second, we declare that~$v_1\wedge\dots\wedge v_r=0$ if~$v_i=v_j$ for
      some~$i\neq j$.  To illustrate these rules in action
      suppose~$u,v,w\in V$. Then in~$\bigwedge^3V$, we have the
      following:
      \begin{align*}
        u\wedge(2u+3v+5w)\wedge w&=
        u\wedge(2u)\wedge w+
        u\wedge(3v)\wedge w+
        u\wedge(5w)\wedge w\\
        &=
        2\,u\wedge u\wedge w+
        3\,u\wedge v\wedge w+
        5\,u\wedge w\wedge w\\
        &=
        0+ 3\,u\wedge v\wedge w+ 0\\
        &=
        3\,u\wedge v\wedge w.
      \end{align*}
      Another example, this time in~$\bigwedge^2V$:
      \begin{align*}
        (u+2v)\wedge(u+3v)
        &= u\wedge(u+3v)+(2v)\wedge(u+3v)\\
        &= u\wedge u+u\wedge(3v)+(2v)\wedge u+(2v)\wedge(3v)\\
      &= 0+3\,u\wedge v+2\,v\wedge u+6\,v\wedge v\\
      &= 3\,u\wedge v+2\,v\wedge u+0\\
      &= 3\,u\wedge v+2\,v\wedge u.
      \end{align*}
      It turns out there is a little more we can do to simplify this last example.
      By the second rule, we have~$(u+v)\wedge (u+v)=0$, since in this expression
      we have two copies of the same vector.  But linearly expanding this expression, we
      get
      \begin{align*}
        0
        &=(u+v)\wedge(u+v)\\
        &=u\wedge(u+v)+v\wedge(u+v)\\
        &=u\wedge u+u\wedge v+v\wedge u+v\wedge v\\
        &=0+u\wedge v+v\wedge u+0\\
        &=u\wedge v+v\wedge u.
      \end{align*}
      Thus,~$u\wedge v+v\wedge u=0$.  This means that
      \[
        u\wedge v=-v\wedge u.
      \]
      In fact, in a wedge product of vectors, swapping any two locations negates
      the expression. (The proof is similar to the one we just gave in the case
        of~$r=2$.)  For instance,
      \[
      u\wedge v\wedge w = -u\wedge w\wedge v = w\wedge u\wedge v=-w\wedge v\wedge
      u.
      \]
      Continuing our example from above, we get
      \begin{align*}
        (u+2v)\wedge(u+3v)
        &= \dots\quad\text{(see earlier calculation)}\\
        &= 3\,u\wedge v+2\,v\wedge u\\
        &= 3\,u\wedge v-2\,u\wedge v\\
        &= u\wedge v.
      \end{align*}
      Now for some problems:
      \begin{enumerate}[(a)]
        \item Let~$V=\R^2$, and take two vectors~$u=(a,b)$ and $v=(c,d)$ in $\R^2$.
  	Let~$e_1=(1,0)$ and~$e_2=(0,1)$.  Writing~$u$ and~$v$ as linear
  	combinations of~$e_1$ and~$e_2$, find the number~$k$ in terms
  	of~$a,b,c,d$ such that
  	\[
  	  u\wedge v=k\, e_1\wedge e_2,
  	\]
  	in~$\bigwedge^2V$.
  	What is the relation between ~$k$ and~$\det\left(\begin{array}{cc} a&b\\c&d\end{array}\right)$?
        \item Now let~$V=\R^3$, and take
  	vectors~$u=(u_1,u_2,u_3)$ and~$v=(v_1,v_2,v_3)$ in~$\R^3$.  Writing these
  	vectors as linear combinations of the standard basis
  	vectors~$e_1=(1,0,0)$,~$e_2=(0,1,0)$, and~$e_3=(0,0,1)$, find
  	numbers~$p,q,r$ in terms of the~$u_i$ and the~$v_i$ such that
  	\[
  	  u\wedge v = p\,e_2\wedge e_3-q\,e_1\wedge e_3+r\,e_1\wedge e_2.
  	\]
  	(Watch out for the minus sign in front of~$q$.)  Physics students may
  	note a relation with the cross product of two vectors in~$\R^3$.
      \end{enumerate}
\end{prob}
\end{document}
