\documentclass[time,mathptm]{article}

\newcommand{\docid}{Solutions to Practice Problems Exam 1} \input
localmacs


\begin{document}
\doctitle
        
The question are not guaranteed to be correct.

(3) (True/False questions)

a.  False. An example is obtained by taking $\vec{u} = \vec{v}$ and
$\vec{w} = -\vec{v}$. 

b.  False. An example is: $A = B = I$ (identity matrix).
 
c. False. The set doesn't contain the zero vector.

d. True. $\cos(t)$ is not a multiple of $\sin(t)$ (nor vice versa).

e. True.  Any spanning set contains a linearly independent spanning set.

f. True.  $Nul(A) = 0$ means that the columns are linearly independent.


(4) a. $H$ = Span$(
\begin{bmatrix}
  1 \\
  1 \\
  0 \\
  0
\end{bmatrix}, 
\begin{bmatrix}
  1 \\
  0 \\
  -2 \\
   1
\end{bmatrix})$.

b.  The solutions are $H_1 = \begin{bmatrix}
  2 \\
  0 \\
  2 \\
  0
\end{bmatrix} + \vec{u}$ and $H_2 = \begin{bmatrix}
  15 \\
  0 \\
  2 \\
  0
\end{bmatrix} + \vec{u}$, and $A = [ \vec{v}_2, \vec{v}_1]$, 
where $\vec{u} \in H$ and $\vec{v}_i \in H_i$.

(5) a.  $T(\begin{bmatrix} a \\ b \\ c \end{bmatrix}) = 
\begin{bmatrix} a \\ b \\ c \\ 0 \end{bmatrix}$.

b. $T(\begin{bmatrix} a \\ b \\ c \end{bmatrix}) = 
\begin{bmatrix} a \\ b \end{bmatrix}$.

c.  
$S = {\begin{bmatrix}
  1 \\ 0 \\ 0 \\ 0
\end{bmatrix}, 
\begin{bmatrix}
  0 \\ 1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
  0 \\ 0 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
  1 \\ 1 \\ 1 \\ 0 
\end{bmatrix}}$.

d.  $\vec{e}_3 + 
\begin{bmatrix}
  1 \\ 1 \\ 1
\end{bmatrix}$.

(6)  a. False.  $I0 = 0$. 

b. False $-I + I$ is not invertible.

c.  True.  

d.  True.

f. False.  $A = 0$.


(7) a. $-KK = I$.  So $(-K)K=I$.  Hence by the invertible matrix
theorem $K$ is invertible.

b.  Just multiple it out.
 
c.  $KK = PJP^{-1}PJP^{-1} = PJ^2P^{-1} = P(-I)P^{-1} = - PP^{-1} =
-I$.

d.  There are infinitely many invertible matrices $P$.


(8) a.  
$\begin{bmatrix}
  0 & \frac{1}{3} & 0 & 0\\
  \frac{1}{2} & 0 & 0 &0 \\
  0 & 0 & 0 & 4\\
  0 & 0 & -1 & 0
\end{bmatrix}$.

b.  
$\begin{bmatrix}
  1 & 1& 9 - a \\
  0 & \frac{1}{3} & 2\\
  0 & 0 & 1
\end{bmatrix}$.

\newpage

c. $
\begin{bmatrix}
  \cos(\theta) & -\sin(\theta) \\
  \sin(\theta) & \cos(\theta)
\end{bmatrix}$.

(9) a. No

b.  Yes $x_1 = -2, x_2 = 3$.


(10) a.  $a_1a_2a_3$.

b.  $6$.


(11) $E =
\begin{bmatrix}
  \sqrt[4]{3} & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
\end{bmatrix}
$.


(12) det$(cI - A) = c^3 - 2c$. So $cI - A$ is invertible unless
$c=0,\sqrt(2),-\sqrt(2)$.



(13) a.  Note that $A$ is row equivalent to the reduced echelon matrix
$B= \begin{bmatrix}
   1 & 0 & 1 & 0\\ 0 & 1 & 3 & 0 \\ ) & 0 & 0 & 1
\end{bmatrix}$
There is one free variable ($x_3$ is free). So $
\begin{bmatrix}
  -1 \\ -3 \\ 1 \\ 0
\end{bmatrix}$ is a basis.

b. The pivots in $B$ occur in the 1st, 2nd and 4th columns. Thus
columns $1,2,4$ of $A$ are pivot columns, hence those columns are a basis.

(14) $A
\begin{bmatrix}
  1 & 1 & 1 \\ 2 & 0 & -1 \\ 0 & 3 & 4
\end{bmatrix}$.


(15) Let $\vec{u}$ be a non-zero vector in $V$, and let $c$ be a
scalar.  Assume that $c\vec u = \vec 0$.  If $ \neq 0$, then we may
speak of the number $\frac{1}{c} = c^{-1}$. Moreover, using properties
of vector spaces, we see that
$$ \vec 0 = c\vec u$$
$$ c^{-1}(\vec 0 = c\vec u)$$
$$
0 = \vec{u}$$
But $\vec u$ is not $\vec 0$, so the only possibility
is that in fact $c = 0$.

(16) a.  $T(g(t)) =
\begin{bmatrix}
  g(0) \\ g(1)
\end{bmatrix}$ and $T(h(t)) = 
\begin{bmatrix}
  h(0) \\ h(1)
\end{bmatrix}$.  Then $T((g+h)(t)) = 
\begin{bmatrix}
  g(0)+ h(0) \\ g(1) + h(1)
\end{bmatrix} = T(g(t)+ T(h(t))$.  $T((cg)(t)) = \begin{bmatrix}
  cg(0) \\ cg(1)
\end{bmatrix} = cT(g(t))$.

b. $f \in \text{Ker}(T)$ iff $f(0) = 0$ and $f(1) = 0$.  Hence $f(t) = c
t(t-1)$. So $f(t) = t(t-1) = t^2 -t$ is a basis.

c.  $T(1-t)= 
\begin{bmatrix}
  1 \\0
\end{bmatrix}$ and $T(t) = 
\begin{bmatrix}
  0 \\1 
\end{bmatrix}$. $ \begin{bmatrix}
  1 \\0
\end{bmatrix}, \begin{bmatrix}
  0 \\1 
\end{bmatrix}$ is a basis for $R^2$.  Hence the range is all of  $R^2$.



(17) Observe that A is row equiv to the following reduced echelon
matrix 
$\begin{bmatrix}
   1 & 0 & 0 & -1 \\
   0 & 1 & 0 & 1 \\
   0 & 0 & 1 & 0  
\end{bmatrix}$.
This reduced echelon matrix has pivots in columns 1,2, and 3, hence
columns 1,2,3 of $A$ form a basis for $H$.

(18) Since the two elements of B are not multiples of one another,
they are linearly independent.  They span $R^2$ and hence they form a
basis of $R^2$.

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