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\noindent
{\bf Solutions to HW 4.}

\bigskip
  
\begin{enumerate}
\item[1.] Describe a way to view the set $\mathbb C$
of complex numbers as a real vector space.
(This means: describe the additive structure
and how to scale elements by a real number.)
\end{enumerate}

\bigskip

\color{blue}
{\bf Solutions.}
According to the usual construction of the complex
numbers, each $z\in\mathbb C$ may be written
as $z=a+bi$ where $a, b\in \mathbb R$ and $i^2 = -1$.
%(That is, $z = (\textrm{real part})+(\textrm{imaginary part})i$.)
We must explain how to add these objects and scale
them by real numbers. We choose the following definitions.
Assume that $z=a+bi, w=c+di\in\mathbb C$ and $r\in \mathbb R$.
\begin{enumerate}
\item[(a)]  (Addition) $z+w=(a+bi)+(c+di) = (a+c) + (b+d)i$.
%  (Real part of sum is sum of real parts of summands,
%  imaginary part of sum is sum of imaginary part of summands.)
\item[(b)]  (Scaling) $r\cdot z = r\cdot (a+bi) = (ra)+(rb)i$.
\end{enumerate}
\medskip

\color{purple}
That's the end of the problem, but let's say more.
It is natural to ask: do these choices for addition and scaling
satisfy the definition
of a ``vector space''? The answer is Yes, but to verify this there
are 12 laws to check, so 12 proofs are required.
That is too much work for a HW problem, but there is a way
to ``cheat'', or ``avoid all that work''.

Call $z=a+bi$ the ``ordinary representation'' of the complex
number $z$ in terms of the real numbers $a$ and $b$. Introduce
a ``column representation'' by writing $z$ as
$\begin{bmatrix} a\\ b\end{bmatrix}\in \mathbb R^2$.
  We defined the addition and scaling of complex numbers
written in the ordinary representation  
so that, using the column representation,
it agrees with the addition and scaling
of elements of $\mathbb R^2$. We ``already know''
that $\mathbb R^2$ satisfies the 12 laws defining
vector spaces, so the definition we gave must satisfy all 12 laws. 
\color{black}

\bigskip
\bigskip

\begin{enumerate}
\item[2.]
Let $\mathbb R[x]$
be the real vector space of polynomials
in the variable $x$ with real coefficients.
Let $S\subseteq \mathbb R[x]$ be the subset
of polynomials whose roots are real numbers.
Determine whether $S$ is a subspace of $\mathbb R[x]$.
(Say the answer, then give the reason.)
\end{enumerate}
\bigskip

\color{blue}
{\bf Solution.}
No, $S$ is not a subspace. $S$ is
not closed under addition.
To see this, note that
$f(x)=x^2-2x+1$ and $g(x)=2x$ have only real roots,
but $(f+g)(x)=x^2+1$ has non-real roots.
\color{black}

\bigskip
\bigskip

\begin{enumerate}
\item[3.]
Let $\mathbb R[x]$
be the real vector space of polynomials in the variable
$x$ with real coefficients. Let $Z\subseteq \mathbb R[x]$
be the subset of those polynomials with roots at
$x=0,1,2$. Determine whether $Z$ is a subspace of
$\mathbb R[x]$. (Say the answer, then give the reason.)
\end{enumerate}
\bigskip

\color{blue}
{\bf Solution.}
Yes, $Z=\{f(x)\in \mathbb R[x]\;|\;f(0)=f(1)=f(2)=0\}$ is a subspace
of $\mathbb R[x]$.
To see this, observe that
\begin{enumerate}
\item[(a)]  $Z$ is closed under addition:
  If $f(x), g(x)\in Z$, then 
  $f(0)=f(1)=f(2)=0=g(0)=g(1)=g(2)$. Now $(f+g)(0)=f(0)+g(0)=0+0=0$.
  Similarly, $(f+g)(1)=(f+g)(2)=0$, so $(f+g)(x)\in Z$.
\item[(b)]  $Z$ is closed under scaling:
  Assume that $r\in \mathbb R$. If $f(x)\in Z$, then
$f(0)=f(1)=f(2)=0$, so 
$r\cdot f(0)=r\cdot f(1)=r\cdot f(2)=r\cdot 0=0$, so 
$r\cdot f(x)\in Z$.  
\end{enumerate}
\color{black}

\end{document}
