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\title{MA20008 Algebra 1, 2004, Sheet 7 Solutions}
\author{Geoff Smith, {\tt http://www.bath.ac.uk/$\sim$masgcs}}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Let $V$ be a two dimensional vector space over $\mathbb R$
with basis
${\bf v_1}, {\bf v_2}$. Let $W$ be a two dimensional vector space over
$\mathbb R$
with basis
${\bf w_1}, {\bf w_2}$. Suppose that $\alpha: V \longrightarrow W$
is a linear map which has matrix
\[ \left(\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right)\]
with respect to these bases. Determine the matrix of $\alpha$ with
respect to new bases ${\bf v_1} + 2{\bf v_2}, {\bf v_1} - {\bf v_2}$ of
$V$
and ${\bf w_1} + {\bf w_2}, {\bf w_1} - {\bf w_2}$ of $W$.}
\newline \noindent {\bf Solution\ }
Let ${\bf x_1 } = {\bf v_1} + 2{\bf v_2},\ {\bf x_2 } =
{\bf v_1} - {\bf v_2}$, and let ${\bf y_1 } = {\bf w_1} + {\bf w_2},\
{\bf y_2} = {\bf w_1} - {\bf w_2}.$
Now let $P$ be the matrix of $id_V$ using the new basis in the domain
and the old basis in the codomain. We have $id({\bf x_1}) ={\bf x_1} =
{\bf v_1} + 2{\bf v_2}$
and $id({\bf x_2}) = {\bf v_1} - {\bf v_2}$, Therefore
\[ P = \left(\begin{array}{cc} 1 & 1 \\ 2 & -1\end{array}\right).\]
Let $Q$ be the matrix of $id_W$ using the old  basis in the domain
and the  new basis in the codomain. We have
$id({\bf w_1}) =  {\bf w_1} = \frac{1}{2}{\bf y_1} + \frac{1}{2}{\bf y_2}$
and
$id({\bf w_2}) = {\bf w_2} = \frac{1}{2}{\bf y_1} - \frac{1}{2}{\bf y_2}.$
Therefore
\[ Q = \frac{1}{2}\left(\begin{array}{cc} 1 & 1 \\ 1 &
-1\end{array}\right).\]
Thus the matrix of $\alpha$ with respect to the new bases is
\[ QAP = \left(\begin{array}{cc} 8 & -1 \\ -3 & 0\end{array}\right).\]
\item {\em Suppose that $A$ is an $m \times n$ matrix with entries in a
field
$F$. Show that there are vector spaces $V$ and $W$ over $F$ of dimensions
$n$
and $m$ respectively, a linear transformation $\alpha: V \longrightarrow
W$,
and choices of bases for these spaces so that the matrix of $\alpha$ with
respect
to these bases is $A$.}\newline \noindent {\bf Solution\ }
Let $V$ be a vector space over $F$ of dimension $n$.
For example we might have $V = F^n$.
Let $W$  be a vector space over $F$ of dimension $m$.
For example we might have $V = F^m$.
Choose ${\bf v_1}, {\bf v_2}, \ldots, {\bf v_n}$ a basis of $V$
and ${\bf w_1}, {\bf w_2}, \ldots, {\bf w_m}$ a basis of $W$.
Define a linear map $\alpha: V \longrightarrow W$ by
stipulating that $\alpha( {\bf v_j}) = \sum_{i=1}^m a_{ij}{\bf w_i}$ for
every $j$. and extending to all of $V$ by linearity. The matrix of the
linear map $\alpha$ is $A = (a_{ij})$ by design.
\item Let $V$ and $W$ be vector spaces over $F$ of dimensions $n$
and $m$ respectively. Show that there is a choice of bases for
$V$ and $W$ such that the matrix of $\alpha$ with respect to these bases
is \[ \left(\begin{array}{ll} I_r & 0_{r, n-r}\\ 0_{m-r, r} &
0_{m-r,n-r}\end{array}\right).\]
\noindent {\bf Solution\ }
Choose a basis for ${\bf v_{r+1}}, {\bf v_{r+2}}, \ldots, {\bf v_n}$ for
$\mbox{Ker }\alpha$
and extend it if necessary to a basis ${\bf v_{1}}, {\bf v_{2}}, \ldots,
{\bf v_n}$
to form a basis of $V$. Now let ${\bf w_i } = \alpha({\bf v_i })$ for $1
\leq i \leq r$,
a collection of linearly independent vectors (q.v. the proof of the
rank-nullity
theorem). Extend this if necesary to a basis ${\bf w_{1}}, {\bf w_{2}},
\ldots, {\bf w_m}$
of $W$. The matrix of $\alpha$ with respect to these bases has the
required shape.

\item Let $V = \mathbb R^2$ viewed as a vector space over $\mathbb R$, and
thought of as the Euclidean plane. Let $\rho_\theta: V \longrightarrow V$
denote the map which is rotation anticlockwise about the origin through
$\theta$ radians. Choose the standard basis ${\bf e_1}, {\bf e_2}$ for
$V$,
where ${\bf e_1} = (1,0)$ and ${\bf e_2} = (0,1)$.
\begin{enumerate}
\item {\em Show that $\rho_\alpha :  V \longrightarrow V$  is a linear map
for every $\alpha \in \mathbb R$.} \newline \noindent {\bf Solution\ }
Perhaps the easiest way to see this is to endow the vector space
with a extra structure, as regard $V$ as $\mathbb C$, the complex numbers,
so that ${\bf e_1}$ is 1 and ${\bf e_2}$ is $i$.
Now anti-clockwise rotation through $\theta$ about the origin
is accomplished by multiplying by $e^{i\theta}$. Now for ${\bf w}$, ${\bf
z}
\in \mathbb C$ and $\lambda, \mu \in \mathbb R$ we have
\[ e^{i\theta}( \lambda {\bf w} + \mu {\bf z})
= \lambda (e^{i\theta}){\bf w}+ \mu(e^{i\theta}){\bf z}\]
which is exactly what we want.

\item {\em Show that $\rho_\alpha \circ \rho_\beta = \rho_{\alpha +
\beta}$
for all $\alpha, \beta \in \mathbb R$.}\newline \noindent {\bf Solution\ }
This is easy now that we are using $\mathbb C$, because this statement
amounts to $e^{i\pi(\alpha + \beta)} =e^{i\pi\alpha}e^{i\pi\beta}$
\item {\em Show that $\rho_\alpha$ and $\rho_\beta$
commute for all $\alpha, \beta \in \mathbb R$.}  \newline
\noindent {\bf Solution\ }
This statement
amounts to $e^{i\pi(\alpha + \beta)} =e^{i\pi(\beta + \alpha)}.$
{\em \item Write down the matrix of
$\rho_\alpha$ with respect to the given basis.}
\newline
\noindent {\bf Solution\ }
\[ \left( \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta
& \cos \theta \end{array}\right).\]
\item {\em Deduce formulas
for $\cos(\alpha + \beta)$ and $\sin(\alpha + \beta).$}
\[  \left( \begin{array}{cc} \cos \alpha & \sin \alpha \\ - \sin \alpha
& \cos \alpha \end{array}\right)
 \left( \begin{array}{cc} \cos \beta & \sin \beta \\ - \sin \beta
& \cos \beta \end{array}\right)
\]\[=
 \left( \begin{array}{cc} \cos (\alpha + \beta) & \sin (\alpha + \beta)\\
- \sin (\alpha + \beta)
& \cos (\alpha + \beta) \end{array}\right).\]
Equating the entries yields the standard formulas for
$\cos (\alpha + \beta)$ and $\sin (\alpha + \beta).$
\item {\em Let $c$ denote complex conjugation. Calculate the matrix of
$c \circ \rho_\theta \circ c$ and comment on what this means.}
Using the standard basis, the matrix of the map defined by complex
conjugation is
\[ \left( \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right).\]
Therefore the required composition is
\[ \left( \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right)
\left( \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta
& \cos \theta \end{array}\right)
\left( \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right) \]\[=
\left( \begin{array}{cc} \cos (-\theta) & \sin (-\theta) \\ - \sin
(-\theta)
& \cos (-\theta) \end{array}\right).\]
Thus $c \circ \rho_\theta \circ c = \rho_{-\theta}.$
\end{enumerate}
\item {\em Suppose that $A$ and $B$ are invertible $n \times n$ matrices
with real entries.
Show that there are invertible $n \times n$ real matrices $D$ and $E$ such
that
$DAE = B$.} Pick a basis for $V$, and use $A$ and this basis
to define a linear map $\alpha$. Now change basis in domain and
domain (as in Question 3) so there are $n \times n$ invertible
matrices $P$ and $Q$ so that $QAP = I_n$. You get the identity matrix
because $\alpha$ is bijective. Similarly there are invertible
$n \times n$ matrices $R$ and $S$ such that $RAS = I_n$.
Now $R^{-1}PAQS^{-1} = B$. Put $D = R^{-1}P$ and $E = QS^{-1}.$
We are done.
\item {\em Define a relation on $n \times n$ matrices with real entries
as follows: if $A, B \in \mbox{Mat}_{n,n}(\mathbb R),$ we write $A \sim B$
if and only if there are invertible $P, Q \in \mbox{Mat}_{n,n}(\mathbb R)$
such that $QAP = B$.}
\begin{enumerate}
\item {\em Show that $\sim$ is an equivalence relation.} \newline
\noindent {\bf Solution\ }
Let $I$ denote the $n \times n$ identity matrix.
Now if $A \in \mbox{Mat}_{n,n}(\mathbb R),$ then
$A = IAI$ so reflexivity is established.
Next suppose that $A, B \in \mbox{Mat}_{n,n}(\mathbb R)$i
and there are invertible $P, Q \in \mbox{Mat}_{n,n}(\mathbb R)$
such that $QAP = B$. Then $Q^{-1}BP^{-1} = A$ so symmetry is
establisheid. Finally
suppose that $A \sim B$ and $B \sim C$. Thus there
are invertible matrices $P_1, P_2, Q_1, Q_2
\in \mbox{Mat}_{n,n}(\mathbb R)$ such that
$P_1AQ_1 = B$ and $P_2BQ_2 = C$.
Therefore
$(P_2P_1)A(Q_1Q_2) = C$ and both $P_2P_1$ and
$Q_1Q_2$ are invertible elements of
$\mbox{Mat}_{n,n}(\mathbb R)$. Transitivity is established.
\item {\em How may equivalence classes are there?}
\newline
\noindent {\bf Solution\ }
There are $n+1$ equivalence classes. Every matrix is equivalent
to one of the form
\[ \left(\begin{array}{ll} I_r & 0_{r, n-r}\\ 0_{m-r, r} &
0_{m-r,n-r}\end{array}\right).\] As explained in lectures, the quantity
$r$ is the rank of any linear transformation giving rise
to the matrix $A$, and is both the row rank and column rank
of $A$.  Therefore as $r$ varies the matrices
\[ \left(\begin{array}{ll} I_r & 0_{r, n-r}\\ 0_{m-r, r} &
0_{m-r,n-r}\end{array}\right)\] are pairwise inequivalent.
\end{enumerate}
\end{enumerate}
\end{document}


