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\usepackage{amssymb,latexsym}                                                                
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{\bf MATH 221 Quiz IV Solutions}
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\bigskip
\bigskip
\noindent
{\bf 1.} \;Let $V_n = \{a_0 + a_1 t + a_2 t^2 + ... + a_n
t^n\,|\,a_i \in {\bf R}, i = 0 , ..., n\}$ be the vector
space of polynomials of degree at most $n$. The set $\{1, t,
t^2, ..., t^n\}$ is a basis for $V_n$ and is referred to as
the standard basis. Consider the case $n = 3$ and show that:

\medskip
({\bf a})
the set $\mathcal{B} = \{1, (1 - t), (1 - t)^2, (1 - t)^3\}$
is also a basis of $V_3$;

\medskip
Solution: {\it Suppose that $c_0 + c_1 (1 - t) + c_2 (1 -
t)^2 + c_3 (1 - t)^3 = 0$ then setting $t = 1$ yields $c_0 =
0$ thus $c_1 (1 - t) + c_2 (1 - t)^2 +
c_3 (1 - t)^3 = 0$. Factoring out the common factor $(1 -
t)$ yields $c_1 + c_2 (1 - t) +
c_3 (1 - t)^2 = 0$. Setting $t = 1$ again yields $c_1 = 0$
and hence $c_2 (1 - t) +
c_3 (1 - t)^2 = 0$. Factoring out $(1 - t)$ gives $c_2
+ c_3 (1 - t) = 0$ and setting $t = 1$ once more shows that
$c_2 = 0$. We arrive at $c_3 (1 - t)^2 = 0$ which implies
that
$c_3 = 0$.}

\medskip
({\bf b}) what are the coordinates of the polynomial $1 + t
+ t^2 + t^3$ relative to the standard basis? 

\medskip
Solution: {\it $[1, 1, 1, 1]$.}

\medskip
({\bf c}) a polynomial $P$ has coordinates $[1, 1, 1,
1]_{\mathcal{B}}$ relative to $\mathcal{B}$ what are the
coordinates of $P$ relative to the standard basis?

\medskip
Solution: {\it $1 = [1, 0, 0, 0], 1 - t = [1, -1, 0, 0], (1
- t)^2 = 1 - 2t + t^2 = [1, -2, 1, 0], (1 - t)^3 = 1 - 3t
+ 3 t^2 - t^3 = [1, -3, 3, -1]$. Writing these as column
vectors results in a matrix
$$P_{\mathcal{B}} = \begin{pmatrix}
1&1&1&1\\
0&-1&-2&-3&\\
0&0&1&3\\
0&0&0&-1\end{pmatrix}$$and
$$P_{\mathcal{B}}\begin{pmatrix}
1\\
1\\
1\\
1\end{pmatrix}_{\mathcal{B}} = \begin{pmatrix}
1&1&1&1\\
0&-1&-2&-3&\\
0&0&1&3\\
0&0&0&-1\end{pmatrix}\begin{pmatrix}
1\\
1\\
1\\
1\end{pmatrix} = \begin{pmatrix}
4\\
-6\\
4\\
-1\end{pmatrix}.$$Thus the coordinates relative to the
standard basis is $[4, -6, 4, -1]$, i.e., it is the
polynomial $4 - 6t + 4t^2 - t^3$.}

\medskip
({\bf d}) what are the coordinates of the polynomial $1 + t
+ t^2 + t^3$ relative to the basis $\mathcal{B}$ given in
$({\bf a})$? 

\medskip
Solution: {\it We use the formula $[{\bf x}]_{\mathcal{B}} =
P_{\mathcal{B}}^{-1} [{\bf x}]$. It is clear that
$P_{\mathcal{B}}^{-1} = P_{\mathcal{B}}$ hence
$$[{\bf x}]_{\mathcal{B}} =
P_{\mathcal{B}}^{-1} [{\bf x}] = \begin{pmatrix}
1&1&1&1\\
0&-1&-2&-3&\\
0&0&1&3\\
0&0&0&-1\end{pmatrix}\begin{pmatrix}
1\\
1\\
1\\
1\end{pmatrix} = \begin{pmatrix}
4\\
-6\\
4\\
-1\end{pmatrix}_{\mathcal{B}},$$i.e., $1 + t + t^2 + t^3 = 4
- 6(1 -t) + 4(1 - t)^2 - (1 - t)^3.$}

\medskip
({\bf e}) the set $\mathcal{C} = \{1, (1 + t), (1 + t)^2, (1
+ t)^3\}$ is also a basis for $V_3$ what is the
coordinates relative to $\mathcal{C}$ of the polynomial $P
= [1, 1, 1, 1]_{\mathcal{B}}$ of part (c)? 

\medskip
Solution: {\it $1 = [1, 0, 0, 0], 1 + t = [1, 1, 0, 0], (1
+ t)^2 = 1 + 2t + t^2 = [1, 2, 1, 0], (1 + t)^3 = 1 + 3t
+ 3 t^2 + t^3 = [1, 3, 3, 1]$. Writing these as column
vectors results in a matrix
$$P_{\mathcal{C}} = \begin{pmatrix}
1&1&1&1\\
0&1&2&3&\\
0&0&1&3\\
0&0&0&1\end{pmatrix},\;P_{\mathcal{C}}^{-1} = \begin{pmatrix}
1&-1&1&-1\\
0&1&-2&3&\\
0&0&1&-3\\
0&0&0&1\end{pmatrix}.$$We know that
$$[{\bf x}]_{\mathcal{C}} = P_{\mathcal{C} \leftarrow
\mathcal{B}} \begin{pmatrix}
1\\
1\\
1\\
1\end{pmatrix}_{\mathcal{B}} = P_{\mathcal{C}}^{-1}
P_{\mathcal{B}} \begin{pmatrix}
1\\
1\\
1\\
1\end{pmatrix}_{\mathcal{B}}$$because
$P_{\mathcal{C} \leftarrow
\mathcal{B}} = P_{\mathcal{C}}^{-1} P_{\mathcal{B}}$ and
$$P_{\mathcal{C}}^{-1}
P_{\mathcal{B}} \begin{pmatrix}
1\\
1\\
1\\
1\end{pmatrix}_{\mathcal{B}} = \begin{pmatrix}
1&-1&1&-1\\
0&1&-2&3&\\
0&0&1&-3\\
0&0&0&1\end{pmatrix} 
\begin{pmatrix}
4\\
-6\\
4\\
-1\end{pmatrix} = \begin{pmatrix}
15\\
-17\\
7\\
-1\end{pmatrix}_{\mathcal{C}}$$}

\bigskip
\bigskip
\noindent
{\bf 2.} \;Given a polynomial $P(t) = a_0 + a_1 t + a_2
t^2 + ... + a_n t^n$ of degree $n$ then its derivative
$P^{'}(t) = a_1 + 2a_2 t + ... + n a_n t^{n-1}$ is a
polynomial of degree at most $n - 1$. Consider the case $n
= 3$ and show that:

\medskip
({\bf a})
the map $T : V_3 \to V_2$ defined by $T(P(t)) = P^{'}(t)$
is a linear transformation;

\medskip
Solution: {\it If $P(t)$ and
$Q(t)$ are two polynomials
of degree at most $3$ then $P(t) + Q(t)$ is also a
polynomial of degree $3$ then $T(P(t) + Q(t)) = (P(t) +
Q(t))^{'} = P^{'}(t) + Q^{'}(t) = T(P(t)) + T(Q(t))$. If $c$
is a scalar then $T(cP(t)) = (c P(t))^{'} = c P^{'}(t) = c
T(P(t))$.}

\medskip
({\bf b}) write down the matrix $A$ representating the
transformation $T$ relative to the standard bases for $V_3$
and $V_2$? 

\medskip
Solution: {\it The standard basis for $V_3 = \{e_1 = 1, e_2
= t, e_3 = t^2, e_4 = t^3\}$ and thot of $V_2 = \{e_1 = 1,
e_2 = t, e_3 = t^2\}$. We have
\begin{align*}Te_1 &= (1)^{'} = 0 = 0(1) + 0(t) + 0(t^2) =
[0, 0, 0]\\
Te_2 &= (t)^{'} = 1 = 1(1) + 0(t) + 0(t^2) = [1, 0, 0]\\
Te_3 &= (t^2)^{'} = 2t = 0(1) + 2(t) + 0(t^2) = [0, 2, 0]\\
Te_4 &= (t^3)^{'} = 3t^2 = 0(1) + 0(t) + 3(t^2) = [0, 0,
3]\end{align*}thus the matrix $A$ is given by
$$A = \begin{pmatrix}0&1&0&0\\
0&0&2&0\\
0&0&0&3\end{pmatrix}$$}

\medskip
({\bf c}) what is the null space of the matrix $A$ in
$({\bf b})$? 

\medskip
Solution: {\it Let ${\bf x} = [x_1, x_2, x-3, x_4]$ be a
vector in the null space of $A$ then
$A{\bf x} = 0$ which is equivalent to the condition that
$x_2 = x_3 = x_4 = 0$ with $x_1$ being a free variable so:
$${\rm nul}\, A = \{x_1 \begin{pmatrix}1\\
0\\
0\\0\end{pmatrix}\}$$}

\medskip
({\bf d}) what is the column space of the matrix $A$ in
$({\bf b})$?

\medskip
Solution: {\it The column space is the linear span of the
vectors 
$$\{\begin{pmatrix}1\\
0\\
0\end{pmatrix}, \begin{pmatrix}0\\
2\\
0\end{pmatrix}, \begin{pmatrix}0\\
0\\
3\end{pmatrix}\}$$which is all of $V_2$.}

\medskip
({\bf e}) verify the Rank Theorem by using your answers for
$({\bf c})$ and $({\bf d})$.

\medskip
Solution: {\it By $(c)$ the nullity of A = $\dim\, {\rm
nul}\, A = 1$ and by $(d)$ the rank of $A = \dim {\rm
column\;space} = 3$ thus nullity of $A$ + rank of $A$ = 4 =
number of columns of $A$ verifying the Rank Theorem.}


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