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\centerline{Mathematics 244 -- Linear Algebra}
\centerline{Solutions -- Makeup Exam 2}
\centerline{April 2, 2004}
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\noindent
I.  See class notes or text.
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\noindent
II.  A) Let $(a_1,a_2,a_3,a_4), (b_1,b_2,b_3,b_4) \in {\bf R}^4$ and 
$c\in {\bf R}$.  Then
$$\eqalign{&T(c(a_1,a_2,a_3,a_4)+(b_1,b_2,b_3,b_4))\cr
&=T(ca_1+b_1,ca_2+b_2,ca_3+b_3,ca_4+b_4)\cr
&=((ca_1+b_1)+(ca_2+b_2)-(ca_4+b_4) + ((ca_2+b_2)-3(ca_3+b_3))x\cr
&\quad + (-(ca_1+b_1) -2(ca_2+b_2)+3(ca_3+b_3)+(ca_4+b_4))x^2\cr
&=c\left((a_1 + a_2 - a_4) + (a_2 - 3a_3)x + 
(-a_1 - 2a_2 + 3a_3 + a_4)x^2\right) \cr
&\quad + (b_1 + b_2 - b_4) + (b_2 - 3b_3)x + 
(-b_1 - 2b_2 + 3b_3 + b_4)x^2\cr
&=cT(a_1,a_2,a_3,a_4)+T(b_1,b_2,b_3,b_4)\cr}$$
Hence $T$ is linear.
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\noindent
B) We have 
$$\eqalign{
T(1,0,0,0) &= 1-x^2 = 0(1+x) + 0(x) + (-1)(x^2-1)\cr
T(0,1,0,0) &= 1+x-2x^2 = (-1)(1+x)+2x + (-2)(x^2-1)\cr
T(0,0,1,0) &= -3x + 3x^2 = 3(1+x) + (-6)x + 3(x^2-1)\cr
T(0,0,0,1) &= -1 + x^2 = 0(1+x) + 0(x) + 1(x^2-1)\cr}$$
These expansions are found by solving systems of linear equations.
For instance:  If 
$$-3x + 3x^2 = a_1(1+x) + a_2x + a_3(x^2-1) = (a_1-a_3) + (a_1+a_2)x+a_3x^2,$$
then equating coefficients of the powers of $x$ on both sides of the
equations: $a_1 - a_3 = 0, a_1 + a_2 = -3, a_3 = 3$.  So
$a_1 = 3$, and $a_2 = -6$, which shows how 
$T(0,0,1,0)$ is derived.  Therefore the matrix
is 
$$[T]_\alpha^\beta = \pmatrix{0&-1&3&0\cr
                              0&2&-6&0\cr
                             -1&-2&3&1\cr}$$
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\noindent
C)  From the formula given for $T$ above (the matrix could also be used
for this, but I'll do it this way):  ${\rm Ker(T)}$ is the 
set of all $(a_1,a_2,a_3,a_4) \in {\bf R}^4$ such that
$T(a_1,a_2,a_3,a_4) = $ the zero polynomial, or equivalently
$$\matrix{&a_1 &+ &a_2 &  &    &- &a_4 &= 0\cr
          &    &  &a_2 &- &3a_3&  &    &= 0\cr
          &-a_1&- &2a_2&+ &3a_3&+ &a_4 &= 0\cr}$$
The equivalent echelon form system is 
$$\matrix{&a_1 &   &+ &3a_3 &- &a_4 &=0\cr
          &     &a_2&- &3a_3 &  &    &=0\cr}$$
(one equation cancels out completely).  This system has 2 free variables
($a_3,a_4$), so $\dim {\rm Ker}(T) = 2$.  A basis for the kernel is
$$\{(-3,3,1,0),(1,0,0,1)\}$$
(the first comes from $a_3 = 1, a_4 = 0$, the second from 
$a_3 = 0, a_4 = 1$).
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\noindent
D)  The Dimension Theorem says 
$4 = \dim {\rm Ker}(T) + \dim {\rm Im}(T)$.  By part C, 
$\dim {\rm Ker}(T) = 2$, so $\dim {\rm Im}(T) = 2$.  Since
$\dim P_2({\bf R}) = 3$, this says that $T$ is {\it not surjective}.
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\noindent
III. 
\item{A)} {\it False.}  The only way this statement would have
to be true would be if ${\rm dim} {\rm Ker}(T) \le 2$.  But
there are linear mappings for this $V$, $W$ for which 
${\rm Ker}(T)$ is 3 or larger, so a set of three vectors in 
the kernel could be linearly independent.  For instance, 
take 3 linearly independent vectors in the kernel of the zero 
mapping (kernel is all of $V$).
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\item{B)} {\it False.}  Composing the rotations gives a rotation
through the {\it sum} of the two rotation angles.
$2\pi/3 + 4\pi/3 = 2\pi$.  So $ST = R_{2\pi} = I_{{\bf R}^2}$.
Using the standard basis $\beta$, we would have
$$[ST]_\beta^\beta = \pmatrix{1&0\cr 0&1\cr}.$$
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\item{C)} {\it True.}  If $\dim {\rm Ker}(T) = 0$, then 
the only element of the kernel is the zero vector in $V$.
Then if $T(v_1) = T(v_2)$ for some vectors $v_1$ and $v_2$ in $V$,
because $T$ is linear, $T(v_1 - v_2) = 0$.  This says $v_1 - v_2 \in {\rm Ker}(T)$
so $v_1 - v_2 = 0$, which implies $v_1 = v_2$.  Hence $T$
is injective.
\bye
