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\textbf{Exercises on transposes, permutations, spaces}
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\begin{problem} (2.7 \#13.  {\em Introduction to Linear Algebra:} Strang) 
\begin{enumerate}[a)]
\item Find a 3 by 3 permutation matrix with $P^3 = I$ (but not $P=I$).

\item Find a 4 by 4 permutation $\widehat{P}$ with $\widehat{P}^4 \neq I.$
\end{enumerate}
\end{problem}

\solution{
\begin{enumerate}[a)]

\item Let $P$ move the rows in a cycle: the first to the second, the second to the third, and the third to the first. So 
$$P = \threebythree 0 0 1 1 0 0 0 1 0 , P^2 = \threebythree 0 1 0 0 0 1 1 0 0 , \text{ and } P^3 = I. $$

\item Let $\widehat{P}$ be the block diagonal matrix with 1 and $P$ on
  the diagonal; $\widehat{P} = \twobytwo 1 0 0 P$. Since $P^3 = I,$
  also $\widehat{P}^3 = I.$ So $\widehat{P}^4 = \widehat{P} \neq I.$
\end{enumerate}
}

\begin{problem} 
Suppose $A$ is a four by four matrix. How many entries of $A$ can be chosen independently if:
\begin{enumerate}[a)]
\item $A$ is symmetric? 

\item $A$ is {\em skew-symmetric}? ($A^T = - A$) 
\end{enumerate}
\end{problem}

\solution{
\begin{enumerate}[a)]
\item The most general form of a four by four symmetric matrix is:
$$A = \left [ \begin{array}{rrrr} 
a & e & f & g \\
e & b & h & i \\
f & h & c & j \\
g & i & j & d \\  \end{array} \right ]. $$
Therefore \textbf{10} entries can be chosen independently. 

\item The most general form of a four by four skew-symmetric matrix is:
$$ A = \left [ \begin{array}{rrrr} 
0 & -a & -b & -c \\
a & 0 & -d & -e \\
b & d & 0 & -f \\
c & e & f & 0 \\  \end{array} \right ]. $$
Therefore $\textbf{6}$ entries can be chosen independently. 
\end{enumerate}
}

\begin{problem} (3.1 \#18.) True or false (check addition or give a counterexample):
\begin{enumerate}[a)]
\item The symmetric matrices in $M$ (with $A^T = A$) form a subspace. 

\item The skew-symmetric matrices in $M$ (with $A^T = -A$) form a subspace.

\item The unsymmetric matrices in $M$ (with $A^T \neq A$) form a subspace.
\end{enumerate}
\end{problem}

\solution{
\begin{enumerate}[a)]

\item True: $A^T = A$ and $B^T = B$ lead to: $$(A+B)^T = A^T + B^T = A + B, \text{ and } (cA)^T = cA.$$

\item True: $A^T = -A$ and $B^T = -B$ lead to: $$(A+B)^T = A^T + B^T = -A - B = -(A+B), \text{ and } (cA)^T = -cA.$$

\item False: $\twobytwo 1 1 0 0 + \twobytwo 0 0 1 1 = \twobytwo 1 1 1 1 .$ 
\end{enumerate}
}

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