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\fancyhead[L]{\Large \textbf{Math 207 Fall 2014}}
\fancyhead[R]{\large \textbf{Problem Set 1}}


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\section*{Correspondences obtained by using ordered bases} 
\subsection*{Using an ordered basis to represent a vector as columns of scalars}
Using an ordered basis $B=\{b_1, \ldots, b_n\}$ for a vector space
$V$, there is a bijective correspondence between the vectors in $V$
and vectors in $\R^n$.  I will remind you of this correspondence and
also define some convenient terminology.

Any vector $v\in V$ can be expressed uniquely
as $v=\sum_{i=1}^n \alpha_i b_i$ where $\alpha_i \in \R$.  Then
$v$ corresponds to the column vector $\Rep_B(v)\in \R^n$ defined
by $$\Rep_B(p(x))=\begin{pmatrix}\alpha_1\\\alpha_2\\\vdots \\\alpha_n\end{pmatrix}.$$
The correspondence 
\begin{align*}
V &\longleftrightarrow \R^n\\
v&\longleftrightarrow \Rep_B(v)
\end{align*}
is one-to-one since every vector in the space can be expressed as a linear
combination of the basis vectors uniquely and the corrspondence is
onto since every linear combination of the basis
vectors is a vector in the space.

Note that the correspondence depends on the choice of ordered basis
$B$, and that dependence is reflected in the notation $\Rep_B(v).$

\bp Let $V$ be the space of polynomials of degree less than or equal
to four.  The set $$B=\{1+x,1-x,1-2x^2,1+x-x^3,x^2-x^4\}$$ is a basis
for $V$.  Use the ordered basis $B$ to obtain a correspondence between
$V\leftrightarrow \R^5.$

\ben 
\item Give the polynomials in $V$ that correspond to the vectors
  $$\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix},
  \quad \begin{pmatrix}1\\0\\0\\-1\\0\end{pmatrix}, \text{ and }\begin{pmatrix}1\\0\\-2\\-1\\1\end{pmatrix}$$
\item Give the vectors in $\R^5$ that correspond to the
  polynomials $$x^3, \quad 3x^2, \quad 2x^4-7, \text{ and }4x^4-14$$
\een
\ep

\subsection*{Using ordered bases to represent linear transformations as
a matrix of scalars}
Given an ordered basis $B_V=\{v_1,
\ldots, v_k\}$ of a vector space $V$ and an ordered basis $B_W=\{w_1,
\ldots, w_n\}$ of $W$, there's a bijective correspondence between
linear transformations $T:V\to W$ and $n\times k$ matrices of
scalars, defined as follows:

Given a linear transformation $T:V\to W$, define a matrix
$\Rep_{B_V,B_W}(T)$ by setting the $i$-th column to be $\Rep_{B_W}(T(v_i)).$

To put it another way, for each basis vector $v_i\in B_V$,
the vector $T(v_i)$ is an element of $W$ and so it 
can be expressed uniquely in terms of the
basis $B_W$:
$$T(v_i)=\sum_{j=1}^n a_{ij}w_j.$$
The numbers $a_{ij}$ for $i=1, \ldots, k$ and $j=1, \ldots, n$ define
the $n\times k$ matrix $$\Rep_{B_V,B_W}(T)=\begin{pmatrix}a_{11} &
  a_{12} & \cdots & a_{1k} \\ 
a_{21} &  a_{22} & \cdots & a_{2k} \\
\vdots & \vdots &  & \vdots \\ 
a_{n1} &  a_{n2} & \cdots & a_{nk} 
\end{pmatrix}.$$
Notice, the $i$-th column of the
matrix $\Rep_{B_V,B_W}(T)$ is the column of scalars $$\begin{pmatrix}
a_{i1} \\ a_{i2} \\ \vdots \\ a_{in}\end{pmatrix}=\Rep_{B_W}(T(v_i)).$$

\bp As before, let $V$ be the space of polynomials of degree less than or equal
to four and let $$B=\{1+x,1-x,1-2x^2,1+x-x^3,x^2-x^4\}$$ be a basis
for $V$.  The function $T:V \to V$ defined by $T(p)=p'(t)+p(0)$ defines a
  linear transformation.   
\ben
\item Using the correspondence $V
  \longleftrightarrow \R^5$, express $T$ as a matrix $\Rep_{B,B}(T).$
\item Compute the polynomial $T(x^3)$ and compute the following product of a
  matrix with a vector: $$\Rep_{B,B}(T)\begin{pmatrix}1\\0\\0\\-1\\0\end{pmatrix}.$$
\item Let $B'=\{1, t, t^2, t^3, t^4\}$ and compute the matrix
  $\Rep_{B,B'}(T).$
\item Find a basis $B''$ of $V$ so that the matrix $\Rep_{B,B''}(T)$
  is \emph{diagonal} with either $1$ or $0$ on the diagonal.
\een
\ep

\section*{Composition of linear transformations}

\bp This is a variation of problems 28 and 29 in section 2.8 of the Apostol's book.  Let $V$
be the space of all real polynomials.  Define linear transformations by
\begin{align*}
D:V&\to V    & S:V &\to V & T:V &\to V  \\
p(t)& \mapsto p'(t) & p(t) & \mapsto tp (t)&  p(t) & \mapsto tp' (t)
\end{align*}
\ben
\item Let $p(t)=2+3t-t^2+4t^3$  Determine the image of $p$ under $$D,
  S, T, DT, TD, DS, SD, ST, TS, DT-TD, T^2D^2-D^2T^2.$$
\item Find the kernel of $T$, $T-\id$, and $DT-2D$.
\item Does $D$ have a left inverse or a right inverse?
\item Let $S:V \to V$ be defined by $p(t)\mapsto t p(t)$.  Compute
  $DS-SD$ and, more generally, $DS^n-S^nD$.
\een
\ep

\section*{Inner products and best approximation}

\bp Let $V$ be the space of polynomials of degree
less than or equal to three and let 
$$\langle p,q \rangle = p(0)q(0)+p'(0)q'(0)+p(1)q(1)+p'(1)q'(1).$$
\ben
\item Prove that $\langle p,q \rangle$ defines an inner product on
  $V$.  In particular, prove that if $\langle p,p\rangle =0$ then
  $p=0$.  \emph{Hint:} first prove the lemma that if $p$ is a
  polynomial and $a$ is a number for which $p(a)=0$ and $p'(a)=0$,
  then there exists a polynomial $q$ so that $p(x)=(x-a)^2 q(x)$.
\item Compute the angle between the polynomials $2t^3-3t^2 $ and $1$.
\item Use Gram-Schmidt to replace the basis $\{1,t,t^2, t^3\}$ by an
  orthogonormal set.
\item The following polynomials form a very nice basis for $V$.
\begin{gather*}
e_1=t(t-1)^2 \\
e_2=t^2(t-1) \\
e_3=(2 t+1) (t-1)^2 \\
e_4=t^2(1-2 (t-1)) 
\end{gather*}
Projecting an arbitrary differentiable function $f$ onto $V$ yields 
a linear combination of the basis vectors 
$$a_1e_1+a_2e_2+a_3e_3+a_4e_4.$$
Describe (simply!) the coefficients $a_i$ in terms of the function
$f$.  The simplicity of this description explains why $\{e_1, e_2,
e_3, e_4\}$ is such a nice basis for $V$.
\een
\ep


\bp 
Each of the pictures below shows a 
polynomial approximation for $\cos(2t).$
Five of them are degree four polynomials obtained by projection onto the subspace
of quartic polynomials, using one of the following
inner products.  
\begin{align*}
\<f,g\>_1&=f(2)g(2)+f'(2)g'(2)+f''(2)g''(2)+f'''(2)g'''(2)\\
 \<f,g\>_2&=\int_0^4 f(t)g(t)dt \\
 \<f,g\>_3&=f(0)g(0)+f(1)g(1)+f(2)g(2)+f(3)g(3)+f(4)g(4)\\
\<f,g\>_4&=f(0)g(0)+f(2)g(2)+f'(2)g'(2)+f''(2)g''(2)+f(4)g(4)\\
\<f,g\>_5&=f(0)g(0)+f'(0)g'(0)+f(2)g(2)+f(4)g(4)+f'(4)g'(4)\\
\end{align*}
The sixth approximation is obtained by ``gluing
together'' two degree two polynomials defined on $[0,2]$ and $[2,4]$,
each determined by projection using, respectively, the inner products
 \begin{align*}
 \<f,g\>_6&=f(0)g(0)+f(2)g(2)+f'(2)g'(2)\\
\<f,g\>_7&=f(2)g(2)+f'(2)g'(2)+f(4)g(4)\\
\end{align*}


\begin{center}
\includegraphics[scale=0.45]{picture5.pdf}\qquad
\includegraphics[scale=0.45]{picture3.pdf}
\end{center}
\begin{center}
\includegraphics[scale=0.45]{picture1.pdf}\qquad
\includegraphics[scale=0.45]{picture2.pdf}
\end{center}
\begin{center}
\includegraphics[scale=0.45]{picture6.pdf}\qquad
\includegraphics[scale=0.45]{picture4.pdf}
\end{center}

\bigskip

Which pictures go with which approximations?\ep

\end{document}

