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\title{MA20008 Algebra 1, 2004, Sheet 8 Solutions}
\author{Geoff Smith, {\tt http://www.bath.ac.uk/$\sim$masgcs}}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Suppose that $V$ is a finite dimensional vector space over a field
$F$, and that $\alpha: V \longrightarrow V$ is a linear map.
Suppose that ${\bf e_1}, {\bf e_2}, \ldots, {\bf e_n}$ is 
a basis of $V$ such that each subspace $\langle  {\bf e_i} \rangle$ is
$\alpha$-invariant. Prove that the matrix of $\alpha$ with respect to
this basis (in both domain and codomain) is diagonal.} \newline \noindent
{\bf Solution\ } For each $i$ we have $\alpha({\bf e_i}) = \lambda_i {\bf e_i}$
for some scalar $\lambda_i \in F$. It follows that the matrix of
$\alpha$ with respect to this basis is 
\[ \left( \begin{array}{cccccc}
\lambda_1 & 0 & 0 & . & . & .\\
0 & \lambda_2 & 0 & . & . & .\\
0 & 0 & \lambda_3 & . & . & .\\
. & . & . & . & . & .\\
. & . & . & . & \lambda_{n-1} & 0 \\
. & . & . & . & 0 & \lambda_n 
\end{array}\right).
\]

\item {\em We 
consider the ordinary inner product on $\mathbb R2$ or $\mathbb R3$
defined as the ``dot'' or ``scalar'' product.
Let the vertices of triangle $ABC$ have position vectors
${\bf a}, {\bf b}$ and ${\bf c}$ respectively.}

\begin{enumerate}

\item[(a)] {\em Show that the point $G$ with position vector 
$\frac{1}{3}({\bf a}+{\bf b}+{\bf c})$ lies on each median
(a line joining a vertex to the midpoint of the opposite side).   
Conclude that the medians are concurrent (at a point $G$ which is called
the {\em centroid} of $\triangle ABC$).}
\newline \noindent {\bf Solution\ }
The position vector of the point $\frac{2}{3}$ of the way down
the median from $A$ is ${\bf a} + \frac{2}{3}(\frac{1}{2}({\bf b} + {\bf c})
- {\bf a}) = \frac{1}{3}({\bf a}+{\bf b}+{\bf c}).$ By symmetry this
point $G$ is on all three medians, which must therefore be concurrent.

\item[(b)] {\em The {\em altitude} of $\triangle ABC$ through $A$ is the
straight line through $A$ which is perpendicular to $BC$.
Show that a point $P$ is on this altitude
if and only if the 
position vector ${\bf r}$ of $P$ satisfies $({\bf r} - {\bf a}) \cdot
({\bf b} - {\bf c}) = 0$.}
\newline \noindent {\bf Solution\ } The equation specifies that
the lines $AP$ and $BC$ are perpendicular, or that $A = P$. This
is the required altitude.
 

\item[(c)] {\em Now we insist that the origin is the circumcentre $O$ of 
$\triangle ABC$. Thus there is a quantity $R$ such that
\[ {\bf a} \cdot {\bf a} = {\bf b} \cdot {\bf b} ={\bf c} \cdot {\bf c} =
R2.\] Show that the point $H$ with position vector 
${\bf a}+{\bf b}+{\bf c}$ is on all three altitudes. Deduce that
the three altitudes of $\triangle ABC$ are concurrent at $H$.
{\em The point $H$ is called the orthocentre of $\triangle ABC$.}}  
\newline \noindent {\bf Solution\ } $H$ is on the altitude from $A$
because $({\bf a}+{\bf b}+{\bf c}- {\bf a}) \cdot
({\bf b} - {\bf c}) = ({\bf b} + {\bf c}) \cdot ({\bf b} - {\bf c})
= {\bf b} \cdot {\bf b} -{\bf c} \cdot {\bf c} = R2 - R2 = 0.$
By symmetry $H$ is on all three altitudes which are therefore
concurrent.
\item[(d)] {\em Deduce that the three points 
$O, G$ and $H$ are colinear, and that
the distances are such that $|OH| = 3|OG|$. {\em The line through 
$O$ and $H$ is called the Euler line of $\triangle ABC$.}}
\newline \noindent {\bf Solution\ } We know that
$\overline{OH}= 3\overline{OG}$ because 
$\overline{OH} = {\bf a}+{\bf b}+{\bf c}$
and 
$\overline{OG} = \frac{1}{3}({\bf a}+{\bf b}+{\bf c})$.
\item[(e)] {\em Let $L$ be the midpoint of $BC$ and $M$ be the midpoint of $AH$.
Let $N$ be the midpoint of $LM$. Find the position vector of $N$
(with the origin still at $O$).}
\newline \noindent {\bf Solution\ } 
$\overline{ON} = \frac{1}{4}({\bf a}+ {\bf a}+{\bf b}+{\bf c})
+ \frac{1}{4}({\bf b}+{\bf c}) = \frac{1}{2}({\bf a}+{\bf b}+{\bf c}).$  
\item[(f)] {\em Deduce that $N$ is the midpoint of $OH$ so that
$O, G, N, H$ are colinear and the ratios of lengths are
\[ |OG|:|GN|:|NH| = 2:1:3.\]}
\newline \noindent {\bf Solution\ }
This is because
\[ \overline{OG} =  \frac{1}{3}({\bf a}+{\bf b}+{\bf c}),\]
\[ \overline{ON} =  \frac{1}{2}({\bf a}+{\bf b}+{\bf c})\]
and 
\[ \overline{OH} =  {\bf a}+{\bf b}+{\bf c}.\]
\item[(g)] {\em Show that $|LM| = R$.}
$\overline{LM} = \frac{1}{2}({\bf a}+ {\bf a}+{\bf b}+{\bf c})
-  \frac{1}{2}({\bf b}+{\bf c}) = \frac{1}{2}{\bf a}.$
Now the length of $\bf a$ is $R$,
so $|LM| = \frac{1}{2}R.$
\item[(h)] {\em Deduce that the 
circle with centre $N$ and radius $R/2$ goes through
the following nine interesting points: the midpoints of the sides 
of $\triangle ABC$, the feet of the altitudes of $\triangle ABC$ and
the three points which are midway between $H$ and each of the
three vertices $A$, $B$ and $C$. {\em This is the 
`nine-point circle' or `Feuerbach circle'.}}
\newline \noindent {\bf Solution\ }
We have shown that the circle with centre $N$ and radius $\frac{1}{2}R$
has $LM$ as a diameter. Let the line $AL$ intersect the line
$BC$ at $D$. Now $\angle LDM$ is a right angle, so by the
converse of the ``angle in a semicircle'' theorem, this circle
passes through $D$. The position vector of $N$ reveals that
this is the same circle if we cyclically permute $A, B$ and $C$,
and the result follows.
\end{enumerate}
\item {\em Let $ABCD$ be a cyclic quadrilateral. A {\em maltitude} is a straight
line through the midpoint of a side which
is perpendicular 
to the opposite side. Show that the 
four maltitudes are concurrent.}
\newline \noindent {\bf Solution\ }
The maltitude from the midpoint of $AB$ which
is perpendicular to $CD$ has equation
\[ ({\bf r} - \frac{1}{2}({\bf a} + {\bf b})) \cdot
({\bf c} - {\bf d}).\] Now the point $P$ with position vector
\[ \frac{1}{2}({\bf a} + {\bf b} + {\bf c} + {\bf d}).\]
The symmetry of this position vector ensures that this point
is on all four maltitudes.
\item {\em Suppose that $V$ is an inner product space, with inner product
denoted by $\langle \ , \ \rangle$. Suppose that $U$ and $W$ are
subspaces of $V$.}
\begin{enumerate}
\item[(a)] {\em Show that 
$(U + W)^\perp = (U \cup W)^\perp = U^\perp \cap W^\perp$.}
\newline \noindent {\bf Solution\ }
The second equality is a formality. As for the first,
both inclusions are formailities.
\item[(b)] {\em Suppose that $U \leq W$. Show that $W^\perp \leq U^\perp.$}
\newline \noindent {\bf Solution\ }
This is a formality.
\end{enumerate}
\item {\em Suppose that $V$ is an inner product space of dimension
$n$, and ${\bf 0} 
\not = {\bf v} \in V$. Prove that $\dim ( \{ {\bf v } \}^\perp) = n-1$.}
\newline \noindent {\bf Solution\ }
Consider the map $\theta: V \longrightarrow F$ defined
by ${\bf x} \mapsto ({\bf x} , {\bf v} ).$ This is a linear map,
and is not the zero map since $\theta({\bf v}) = ({\bf v},{\bf v}) \not = 0.$
It is linear, and $F$ is 1-dimensional, so $\theta$ is surjective.
The rank-nullity theorem applies and we are done.
\end{enumerate} 
\end{document}


