\magnification=\magstep1
\centerline{Mathematics 244 -- Linear Algebra}
\centerline{Solutions -- Exam 2}
\centerline{March 29, 2004}
\bigskip
\noindent
I.  
\item{A)}  Let $T : P_3({\bf R}) \to {\bf R}^3$ be the mapping
defined by $T(p) = (p(1),p'(1),p''(1))$.  Then for all 
$p,q \in P_3({\bf R})$ and all $c \in {\bf R}$, we have 
by the sum and scalar multiple rules for derivatives:
$$\eqalign{
T(cp+q) &= ((cp+q)(1),(cp+q)'(1),(cp+q)''(1))\cr
&=(cp(1)+q(1),cp'(1)+q'(1),cp''(1)+q''(1))\cr
&=c(p(1),p'(1),p''(1)) + (q(1),q'(1),q''(1))\cr
&=cT(p) + T(q)\cr}$$
So $T$ is linear.
\bigskip
\item{B)}  Let $\alpha=\{1,x+x^2,x^2+x^3,x^3+1\}$.
Since we know $\dim P_3({\bf R}) = 4$ and there
are 4 vectors in $\alpha$, to show $\alpha$ is 
a basis, it suffices to show that $\alpha$ is 
linearly independent.  So if 
$$0=a_1(1) + a_2(x+x^2)+a_3(x^2+x^3)+a_4(x^3+1)$$
then collecting like powers of $x$ on both sides of the 
equation:
$$0 = (a_1+a_4) + a_2 x + (a_2+a_3)x^2 + (a_4+a_3)x^3$$
implies
$$a_1 + a_4 = 0, \quad a_2 = 0, \quad a_2+a_3 = 0, \quad a_4 + a_3 = 0$$
It easily follows that all $a_i = 0$, so $\alpha$ is 
linearly independent.
\bigskip
\item{C)}  To find the matrix of $T$ with respect
to $\alpha$ in the domain and the standard basis 
$\beta = \{e_1, e_2, e_3\}$ in ${\bf R}^3$, we follow
the usual ``recipe'':
$$\eqalign{
T(1) &= (1,0,0) = 1 e_1 + 0 e_2 + 0 e_3\cr
T(x+x^2) &= (x+x^2,1+2x,2)|_{x=1} = (2,3,2) = 2e_1 + 3e_2 + 2e_3\cr
T(x^2+x^3) &= (2,5,8) = 2e_1 + 5e_2 + 8e_3\cr
T(x^3+1) &= (2,3,6) = 2e_1 + 3e_2 + 6e_3\cr}$$
The scalars from these linear combinations go in the 
columns of the matrix:
$$[T]_\alpha^\beta = \pmatrix{1 & 2 & 2 & 2\cr
                              0 & 3 & 5 & 3\cr
                              0 & 2 & 8 & 6\cr}$$
\bigskip
\noindent
D) Following the same procedure as in part C,
$$\eqalign{
S(e_1) &= (3,1,1) = 3e_1 + 1e_2 + 1e_3\cr
S(e_2) &= (-1,1,0) = -1e_1 + 1e_2 + 0e_3\cr
S(e_3) &= (0,1,0) = 0e_1 + 1e_2 + 0e_3\cr}$$
So 
$$[S]_\beta^\beta = \pmatrix{3 & -1 & 0\cr
                             1 & 1 & 1\cr
                             1 & 0 & 0\cr}$$
Hence by our general results about composition of linear mappings
and matrix multiplication,
$$\eqalign{
[ST]_\alpha^\beta &= [S]_\beta^\beta [T]_\alpha^\beta\cr
&= \pmatrix{3 & -1 & 0\cr
            1 & 1 & 1\cr
            1 & 0 & 0\cr}\pmatrix{1 & 2 & 2 & 2\cr
                                  0 & 3 & 5 & 3\cr
                                  0 & 2 & 8 & 6\cr}\cr
&=  \pmatrix{3 & 3 & 1 & 3\cr
             1 & 7 & 15 & 11\cr
             1 & 2 & 2 & 2\cr}\cr}$$
\bigskip
\noindent
II.  See class notes, notes from review session, or 
the text for the statement and the proof of the Dimension
Theorem.
\bigskip
\noindent
III.  
\item{A)}  This is {\it TRUE} (i.e. true for {\it all linear
mappings} $T : {\bf R}^3 \to {\bf R}^5$).  {\it Proof:} 
If $T : {\bf R}^3 \to {\bf R}^5$ is any linear mapping,
then the Dimension Theorem says
$$\dim {\bf R}^3 = 3 = \dim {\rm Ker}(T) + \dim {\rm Im(T)}$$
Since $\dim {\rm Ker}(T) \ge 0$, this says $\dim {\rm Im}(T) \le 3$.
Hence ${\rm Im}(T) \ne {\bf R}^5$, so there are some 
${\bf b} \in {\bf R}^5$ such that there are no solutions
of the equation $T({\bf x}) = {\bf b}$.
\bigskip
\item{B)}  This is {\it FALSE}.  We proved a statement saying
the opposite, but only in the case that $T : V \to W$
with $V,W$ {\it finite-dimensional} of the same dimension.
Since $P({\bf R})$ is not finite-dimensional that result 
does not apply in this case.  However just because our
other result does not apply is not a sufficient reason
to say the statement is false here.  Instead, we need to 
give an explicit counterexample.  First note that
$\{1,x,x^2,x^3,\ldots\ \} = \{x^k : k \ge 0\}$ is a 
basis of $P({\bf R})$.  By one of our general results, 
to define a linear mapping from $P({\bf R})$
to itself, it suffices to say where the elements of 
this basis map.  To get an example of a non-injective
$T$ that {\it is surjective}, we can say (for instance):
$$\eqalign{T(1) &= 1\cr
           T(x) &= 1\cr
           T(x^2) &= x\cr
                  &\vdots\cr
           T(x^k) &= x^{k-1}\cr}$$
(all $k \ge 1$)  There is a unique linear mapping satisfying this.
$T$ is certainly not injective since $T(1) = T(x)$.  (Says that
${\rm Ker}(T)$ contains $p(x) = x - 1$.)  On the other hand,
$T$ is surjective since a basis for $P({\bf R})$ is contained
in ${\it Im}(T)$.  (Note:  Another example of the same type
is given by $T : P({\bf R}) \to P({\bf R})$ where $T(f) = f'$
-- the derivative mapping!)
\bigskip
\item{C)} This is {\it FALSE}.  The three given vectors are 
{\it linearly independent} since if
$$a_1(1,1,1,0,0,0,0) + a_2(0,0,0,1,1,1,1)+a_3(1,0,1,0,1,0,1) = (0,0,0,0,0,0,0)$$
then components 1, 2, 4 give:
$$a_1 + a_3 = 0, a_1 = 0, a_2 = 0,$$
which implies $a_1 = a_2 = a_3 = 0$.  Hence if these vectors
were in the image of $T$ we would have $\dim {\rm Im}(T) \ge 3$.
(You can't tell $\dim {\rm Im}(T) \ge 3$ without checking this.)
But the Dimension Theorem says 
$$4 = \dim {\bf R}^4 = \dim {\rm Ker}(T) + \dim {\rm Im}(T)$$
so if $\dim {\rm Ker}(T) = 2$, then $\dim {\rm Im}(T) = 2$.
\bigskip
\noindent
{\it Extra Credit}
\bigskip
Let the system be 
$$\matrix{ a_{11} x_1 &+ &\cdots &+ &a_{1n} x_n &= 0\cr
               \vdots &  &\ddots &  &\vdots &\vdots \cr
           a_{m1} x_1 &+ &\cdots &+ &a_{mn} x_n &= 0\cr}$$
The $m\times n$ matrix $A = (a_{ij})$ defines the linear
mapping $T_A : {\bf R}^n \to {\bf R}^m$ (where 
$T_A(x) = Ax$).  The solutions of the homogeneous system
are exactly the vectors in ${\rm Ker}(T) \subset {\bf R}^n$.
The image of $T_A$ is contained in ${\rm R}^m$ so has
dimension $\le m$.
Since $m < n$, the Dimension Theorem says
$$\dim {\rm Ker}(T) = n - \dim {\rm Im}(T) \ge n - m > 0$$
Hence ${\rm Ker}(T)$ contains a nonzero vector -- a nontrivial
solution of the system of equations.  

\bye
