\magnification=\magstep1
\overfullrule=0pt
\def\sol{\bigskip\item{}{\it Solution:}\ }
\centerline{Mathematics 244, section 1 -- Linear Algebra}
\centerline{Final Examination Solutions}
\centerline{May 10, 2007}
\bigskip
\noindent
I.  Let 
$$A = \pmatrix{-1 & 0 & 2  & 2 & -1\cr
                2 & 1 & -1 & 2 & -1\cr
                1 & a & -5 &-8 &  a\cr}$$
\item{A)}  (10) Are there any $a \in {\bf R}$ such that
columns 1,2,3 of $A$ form a linearly dependent set?  If so, 
find such an $a$.  If not, say why not.
\sol
Performing row operations on the first three columns:
$$\eqalign{
\pmatrix{-1 & 0 & 2\cr
          2 & 1 & -1\cr
          1 & a & -5\cr}
&\sim 
\pmatrix{-1 & 0 & 2\cr
          0 & 1 & 3\cr
          0 & a & -3\cr}\cr
&\sim
\pmatrix{-1 & 0 & 2\cr
          0 & 1 & 3\cr
          0 & 0 & -3 - 3a\cr}\cr}$$
The first three columns will be linearly dependent if
$-3 - 3a = 0$, or $a = -1$.  They will be linearly independent
for any other $a \in {\bf R}$.
\bigskip
\item{B)} (10) Are there any $a \in {\bf R}$ such that
the system of equations $Ax = e_1$ (the first standard basis vector
in ${\bf R}^3$) is inconsistent?  If so, for which $a$ is this true?
If not, say why not.
\sol
Do the same row operations on the augmented matrix of the system 
$Ax = e_1$:
$$\eqalign{
&\quad\pmatrix{-1 & 0 & 2 & 2 & -1 & | & 1\cr
          2 & 1 & -1 & 2 & -1 & | & 0 \cr
          1 & a & -5 & -8 &  a & | & 0 \cr}\cr
&\sim 
\pmatrix{-1 & 0 & 2 &2 & -1 & | & 1\cr
          0 & 1 & 3 &6 & -3 & | & 2\cr
          0 & a & -3&-6& a-1 & | & 1\cr}\cr
&\sim
\pmatrix{-1 & 0 & 2 & 2 & -1 & | & 1 \cr
          0 & 1 & 3 & 6 & -3 & | & 2\cr
          0 & 0 & -3 - 3a& -6 - 6a & 4a - 1 & | & 1 - a \cr}\cr}$$
From this we can see that the answer to the question is {\it No}.
The first two columns always contain pivots. 
If $a \ne -1$, then there is a third pivot in column 3, so 
there are solutions.  If $a = -1$, then there is a third pivot
in column 5, since $4(-1) - 1 = -5 \ne 0$.  Therefore,
the system $Ax = e_1$ has solutions for all $a \in {\bf R}$.
\bigskip
\item{C)} (15)  Let $a = 4$ and find a parametrization for 
${\rm Nul}(A)$.
\sol
Continuing from above with $a = 4$ (but omitting the last column):
$$\eqalign{
&\quad \pmatrix{-1 & 0 & 2 & 2 & -1 \cr
          0 & 1 & 3 & 6 & -3 \cr
          0 & 0 & -15 & -30 & 15 \cr}\cr
&\sim 
\pmatrix{ 1 & 0 & -2 & -2 & 1  \cr
          0 & 1 & 3 & 6 & -3 \cr
          0 & 0 & 1 & 2 & -1 \cr}\cr
&\sim 
\pmatrix{ 1 & 0 & 0 & 2 & -1 \cr
          0 & 1 & 0 & 0 & 0 \cr
          0 & 0 & 1 & 2 & -1\cr}\cr}$$
Hence
$$Nul(A) = {\rm Span}\left\{\pmatrix{-2\cr 0 \cr -2 \cr 1 \cr 0}, \pmatrix{1\cr 0\cr 1\cr 0 \cr 1}\right\} = 
\left\{s \pmatrix{-2\cr 0 \cr -2 \cr 1 \cr 0} + t\pmatrix{1\cr 0\cr 1\cr 0 \cr 1} : s,t \in {\bf R}\right\} $$
\bigskip
\noindent
II. 
\item{A)} (5)  {\it Define:}  The {\it span} of a set $S$ in a 
vector space $V$.  
\sol
The span of $S$ is the set of all linear combinations of vectors in $S$, or
$${\rm Span}(S) = \{c_1 x_1 + \cdots + c_k x_k : x_i \in S, c_i \in {\bf R}\}.$$
\item{B)} (10) Given:  if $V$ is a vector space, $S$ is 
a finite subset of $V$ with ${\rm Span}(S) = V$, and
$T$ is any linearly independent subset of $V$,
then $|T| \le |S|$.  Use this fact to show that any
two bases of $V$ must have the same number of elements.
\sol
Let $B$ and $C$ be any two bases of $V$.  Since $B$ is linearly independent
and $C$ spans $V$, we have by the given fact $|B| \le |C|$.  On the other hand,
since $B$ spans $V$ and $C$ is linearly independent, we have $|B| \ge |C|$.
Since both inequalities are true, $|B| = |C|$. 
\bigskip
\item{C)} (5) The result of part B
provides the justification for the definition 
of what concept connected to vector spaces? 
\sol
This is the justification for the definition of the {\it dimension} 
of a vector space as the number of vectors in a basis.  
\vfill\eject
\noindent
III. 
\item{A)} (15) Solve by Cramer's Rule:
$$\eqalign{3x_1\ \ \qquad + x_3 &= 3\cr
           4x_1 + 3x_2 - x_3 &= 2\cr
           \qquad  x_2 + 2x_3 &= 1\cr}$$
\sol
Computing determinants, we have 
$$\eqalign{\det(A) &= 3\cdot 7 - 0 + 1 \times 4 = 25\cr
           \det(A_1(b)) &= \det \pmatrix{3& 0 & 1\cr
                                         2& 3 &-1\cr
                                         1& 1 &2 \cr} = 3\times 7 + 1(-1) = 20\cr
           \det(A_2(b)) &= \det \pmatrix{3& 3 & 1\cr
                                       4& 2 &-1\cr
                                       0& 1 &2 \cr} = 3\times 5 - 3\times 8 + 1\times 4 = -5\cr
           \det(A_1(b)) &= \det \pmatrix{3& 0 & 3\cr
                                         4& 3 & 2\cr
                                         0& 1 & 1\cr} = 3\times 1 + 0 + 3\times 4 = 15\cr}$$
Therefore,
$$\eqalign{
x_1 &= {\det(A_1(b))\over \det(A)} = {20\over 25} = {4\over 5}\cr
x_2 &= {\det(A_2(b))\over \det(A)} = {-5\over 25} = -{1\over 5}\cr
x_3 &= {\det(A_3(b))\over \det(A)} = {15\over 25} = {3\over 5}.\cr}$$
(It is easy to check that these are correct, by substituting in the original equations.)
\bigskip
\item{B)} (5)  Suppose $\det(A) = 3$, $\det(B) = -2$, and $\det(C) = 4$.
What is $\det(A^t B^{-1} C^2)$?
\sol
By the properties of determinants,
$$\det(A^t B^{-1} C^2) = \det(A) \det(B)^{-1} (\det(C))^2 = 3\times {1\over (-2)} \times 4^2 = -24.$$
\bigskip
\noindent
IV.  All parts of this problem refer to the linear mapping 
$T : P_3({\bf R}) \to {\bf R}^3$ defined by
$$T(p) = \pmatrix{p(-1)\cr p(0) \cr p(1)\cr}$$
\item{A)} (10) Show that the {\it image} of $T$ is a vector subspace
of ${\bf R}^3$ {\it directly} (that is, do not just quote a general theorem).
\sol
The image ${\rm Im}(T)$ is the set of $y \in {\bf R}^3$ such that $y = T(p)$ for some
polynomial $p \in P_3({\bf R})$.  Since $T(0) = 0$, we know
$0 \in {\rm Im}(T)$.  Let $y_1, y_2 \in {\rm Im}(T)$.
Then $y_1 = T(p_1)$ and $y_2 = T(p_2)$ for some $p_1,p_2$.  But then
since $T$ is linear:
$$y_1 + y_2 = T(p_1) + T(p_2) = T(p_1 + p_2).$$
This shows $y_1 + y_2 \in {\rm Im}(T)$ also.
Finally if $y = T(p) \in {\rm Im}(T)$ and $c\in {\bf R}$, then
$cy = cT(p) = T(cp)$.  Hence $cy \in {\rm Im}(T)$.  Therefore,
${\rm Im}(T)$ is a subspace of ${\bf R}^3$.  
\bigskip
\item{B)} (10)  Find the matrix of $T$ with respect to the 
basis ${\cal B} = \{1,x,x^2,x^3\}$ in the domain and the standard basis in 
${\bf R}^3$.
\sol
We have
$$
T(1) = \pmatrix{1\cr 1\cr 1\cr}, \quad
T(x) = \pmatrix{-1 \cr 0 \cr 1\cr}, \quad 
T(x^2) = \pmatrix{1 \cr 0 \cr 1\cr}, \quad {\rm and}\quad T(x^3) = T(x).$$
Therefore, by definition:
$$[T]_{std \leftarrow {\cal B}} = \pmatrix{1 & -1 & 1 & -1\cr
                                           1 &  0 & 0 & 0\cr
                                           1 &  1 & 1 & 1\cr}.$$
\item{C)} (15)  State the change of coordinates formula and use it 
to find the matrix of $T$ with respect to 
the basis ${\cal B}$ in the domain and the basis
$${\cal C} = \left\{\pmatrix{1\cr 1\cr 0\cr}, \pmatrix{1\cr 0\cr 1\cr},
\pmatrix{0\cr 1 \cr 1\cr}\right\}$$
in ${\bf R}^3$.  For full credit, you must show all details of the computation
of any matrix inverses you need.  
\sol
The change of coordinates formula says that if $T : V \to W$ is linear,
${\cal B}, {\cal B}'$ are bases of $V$, and ${\cal C} {\cal C}'$ are bases of $W$,
then 
$$[T]_{{\cal C}' \leftarrow {\cal B}'} = 
P_{{\cal C}' \leftarrow {\cal C}} 
[T]_{{\cal C} \leftarrow {\cal B}} 
[T]_{{\cal B} \leftarrow {\cal B}'}.$$
Here we want:
$$\eqalign{[T]_{{\cal C} \leftarrow {\cal B}} &= 
P_{{\cal C} \leftarrow std} 
[T]_{std \leftarrow {\cal B}} \cr
&= (P_{std \leftarrow {\cal C}})^{-1} 
[T]_{std \leftarrow {\cal B}} \cr
&= \pmatrix{1 & 1 & 0\cr
            1 & 0 & 1\cr
            0 & 1 & 1\cr}^{-1} \pmatrix{1 & -1 & 1 & -1\cr
                                           1 &  0 & 0 & 0\cr
                                           1 &  1 & 1 & 1\cr}\cr
&= \pmatrix{1/2 & 1/2 & -1/2\cr
            1/2 & -1/2 & 1/2\cr
            -1/2 & 1/2 & 1/2\cr} \pmatrix{1 & -1 & 1 & -1\cr
                                           1 &  0 & 0 & 0\cr
                                           1 &  1 & 1 & 1\cr}\cr
&= \pmatrix{1/2 & -1 & 0 & -1\cr
            1/2 & 0 & 1 & 0\cr
            1/2 & 1 & 0 & 1\cr}
}$$
The inverse matrix is computed as usual by row reducing
$(P | I) \sim (I | P^{-1})$; details omitted.
\bigskip
\noindent
V.  (10) Show by mathematical induction: If $A = Q^{-1}BQ$ for some invertible $Q$, 
then for all integers $n\ge 1$, $A^n = Q^{-1} B^n Q$.
\sol
The base case of the induction is $n = 1$, and there is nothing to prove.
For the induction step, assume $A^k = Q^{-1}B^kQ$.  Then using
the induction hypothesis,
$$A^{k+1} = A\cdot A^k = (Q^{-1}BQ)(Q^{-1}B^kQ)$$
$$ = Q^{-1}B(QQ^{-1})B^kQ = Q^{-1}BB^k Q = Q^{-1}B^{k+1}Q,$$
which is what we wanted to show.
\bigskip
\noindent
VI.  Let $A$ be the symmetric matrix:
$$A = \pmatrix{a & 0 & b\cr
               0 & a & 0\cr
               b & 0 & a\cr}$$
where $a,b$ are real numbers with $b > a > 0$. 
\item{A)} (10) Show that $A$ has eigenvalues $\lambda
= a + b, a, a - b$ and determine the corresponding eigenspaces. 
\sol
We have
$$\det{A - \lambda I} = \det \pmatrix{a - \lambda & 0 & b\cr
                                      0 & a - \lambda & 0\cr
                                      b & 0 & a - \lambda\cr}.$$
Expanding along row 2, we have
$$= (a-\lambda) \det\pmatrix{a - \lambda & b \cr b & a - \lambda\cr} 
= (a -\lambda)(\lambda^2 - 2a \lambda + (a^2 - b^2)).$$
This factors easily as
$$ = (a -\lambda)((a + b) - \lambda)((a - b) - \lambda).$$
Hence the eigenvalues are $\lambda = a, a+b, a-b$.
Under the assumption $b > a > 0$, all three eigenvalues are distinct. 
We have
$$\eqalign{
{\rm Nul}(A - aI) &= {\rm Nul} \pmatrix{0 & 0 & b\cr 0 & 0 & 0\cr b & 0 & 0\cr} 
= {\rm Span}\left\{\pmatrix{0 \cr 1 \cr 0\cr}\right\}\cr
{\rm Nul}(A - (a+b)I) &= {\rm Nul} \pmatrix{-b & 0 & b\cr 0 & -b & 0\cr b & 0 & -b\cr} 
= {\rm Span}\left\{\pmatrix{1/\sqrt{2} \cr 0 \cr 1/\sqrt{2}\cr}\right\}\cr
{\rm Nul}(A - (a-b)I) &= {\rm Nul} \pmatrix{b & 0 & b\cr 0 & b & 0\cr b & 0 & b\cr} 
= {\rm Span}\left\{\pmatrix{-1/\sqrt{2} \cr 0 \cr 1/\sqrt{2}\cr}\right\}\cr}$$
\bigskip  
\item{B)} (10) Find the spectral decomposition of $A$.  
\sol 
Write $u_i$ for the basis vectors above for the three eigenspaces.
Note that we normalized so that $\{u_1,u_2,u_3\}$ is an orthonormal
basis of ${\bf R}^3$ (this is necessary for the next step).
The spectral decomposition is $A = au_1u_1^t + (a + b)u_2u_2^t + (a-b)u_3u_3^t = $
$$ a\cdot \pmatrix{0 & 0 & 0\cr
                      0 & 1 & 0\cr
                      0 & 0 & 0\cr} + 
(a+b) \cdot \pmatrix{1/2 & 0 & 1/2\cr
                     0   & 0 & 0\cr
                     1/2 & 0 & 1/2\cr} + 
(a-b) \cdot \pmatrix{1/2 & 0 & -1/2\cr
                     0   & 0 & 0\cr
                     -1/2 & 0 & 1/2\cr}.$$
\bigskip
\noindent
VII.  True - False.  For each true statement give a brief proof
(saying ``this is true by definition'' is {\it not sufficient}!).  For
each false statement give a counterexample or a reason.
\bigskip
\item{A)} (10) If $A$ is an $n\times n$ matrix
and $AQ = QD$ for some diagonal matrix $D$ and some matrix $Q$,
then $A$ is diagonalizable.  
\sol
This is {\bf False}.  The equation $AQ = QD$ does mean that all the columns of 
$Q$ are eigenvectors of $A$.  But it is not stated that $Q$ is invertible, 
which means that the columns of $Q$ do not need to form a basis
for ${\bf R}^n$, and $A$ does not need to be diagonalizable.
A counterexample:  let $A = \pmatrix{1 & 1\cr 0 & 1\cr}$ which
is not diagonalizable.  If $Q = \pmatrix{1 & 1\cr 0 & 0\cr}$,
then 
$$AQ = \pmatrix{1 & 1\cr 0 & 1\cr}\pmatrix{1 & 1\cr 0 & 0\cr} = \pmatrix{1 & 1\cr 0 & 0\cr} = 
\pmatrix{1 & 1\cr 0 & 0\cr} I_2 = QD,$$
where $D = I_2$ is the identity matrix.
\bigskip
\item{B)} (10) If $A \in M_{4 \times 5}({\bf R})$ and $\dim {\rm Col}(A) = 2$,
then there are $4$ free variables in the system $Ax = 0$.  
\sol
This is {\bf False}.  A reason is that $\dim {\rm Nul}(A) +\dim {\rm Col}(A) = 5$, 
so if $\dim {\rm Col}(A) = 2$, then $\dim {\rm Nul}(A) = 3$.  Since the set
of solutions of $Ax = 0$ is ${\rm Nul}(A)$ by definition, 
there will only be 3 free variables here.
\bigskip
\item{C)} (10) If $\{v_1,v_2,v_3\}$ is a linearly 
independent subset of ${\bf R}^3$ and $V$ is the matrix with columns 
$v_1,v_2,v_3$, then $\det(V) \ne 0$.  
\sol
This is {\bf True}.  By one of the parts of the Invertible Matrix
Theorem (see p. 129 in the text), $V$ is invertible.  Hence $\det(V) \ne 0$.
\bigskip
\item{D)} (10) If $A$ is a symmetric matrix and $v_1, v_2, \ldots, v_k$ are
eigenvectors of $A$ with distinct corresponding eigenvalues, 
then $\{v_1, v_2, \ldots, v_k\}$ is an orthogonal set.
\sol
This is {\bf True}.  Suppose $A v_i = \lambda_i v_i$ with distinct $\lambda_i$.
Then for all pairs $i \ne j$,
$$\eqalign{\lambda_i (v_i \cdot v_j) &= (\lambda_i v_i)\cdot v_j\cr
                                     &= (Av_i) \cdot v_j\cr
                                     &= v_i \cdot (Av_j)\quad A\ {\rm symmetric}\cr
                                     &= v_i \cdot (\lambda_j v_j)\cr
                                     &= \lambda_j(v_i \cdot v_j)\cr}$$
This shows $0 = (\lambda_i - \lambda_j)(v_i \cdot v_j)$.  Since the first
factor is not zero, the second one must be.  Therefore $v_i$ and $v_j$ are
orthogonal.  Since this is true for all pairs $i,j$, the set of eigenvectors
 is an orthogonal set.
\bigskip
\noindent
VIII.  (The Cayley-Hamilton Theorem.)
\bigskip
\item{A)}  (10)  Let $A = \pmatrix{a & b \cr c & d\cr}$ be 
any $2\times 2$ matrix.  Show
that $A$ ``satisfies its own characteristic polynomial equation''
in the sense that $A^2 - {\rm Tr}(A) A + \det(A) I_2 = 0_{2\times 2}$
(here $I_2$ is the $2\times 2$ identity matrix and $0_{2\times 2}$
is the $2\times 2$ zero matrix).
\sol
Let $A = \pmatrix{a & b\cr c & d\cr}$.
By direct calculation: $A^2 - {\rm Tr}(A) A + \det(A) I_2 = $
$$\eqalign{
&\quad \pmatrix{a^2 + bc & ab + bd\cr
        ca + dc & cb + d^2\cr} - (a+d)\pmatrix{a&b\cr c&d\cr} + 
\pmatrix{ad - bc & 0 \cr 0 &ad - bc\cr}\cr
& = \pmatrix{a^2 + bc - a^2 - ad + ad - bc & ab + bd - ab - bd\cr
                  ca + dc - ac - dc & cb + d^2 - ad - d^2 + ad - bc}\cr
&= \pmatrix{0 & 0\cr 0 & 0\cr}.\cr}$$ 
Thus the Cayley-Hamilton theorem is true for $2\times 2$ matrices.
\bigskip                   
\item{B)}  (10) Prove that an $n\times n$ diagonalizable
matrix $A$ satisfies its own characteristic polynomial in the
same sense as in part A:  If 
$$\det(A - tI) = (-1)^n t^n + c_{n-1} t^{n-1} + \cdots + c_1 t + c_0,$$
then 
$$(-1)^n A^n + c_{n-1} A^{n-1} + \cdots + c_1 A + c_0 I_n = 0_{n\times n}.$$
Caution:  It {\it does not} make sense to ``substitute $t = A$'' in
$\det(A - tI)$'' because $t$ is a scalar, not a matrix.
The result of problem V above may be useful.
({\it Comment:} The result is also true without the extra hypothesis of diagonalizability
but it is more difficult to prove then.)
\sol
Under the assumption that $A$ is diagonalizable, there is an 
invertible matrix $Q$ and a diagonal matrix $D$ with diagonal
entries $\lambda_i$, the eigenvalues of $A$, such that $A = Q^{-1}DQ$.
By problem V above, for all positive integers $k$, $A^k = Q^{-1}D^kQ$,
and the diagonal entries of $D^k$ are just the powers of the 
eigenvalues $\lambda_i^k$.  Substituting for 
the various powers of $A$, and factoring out $Q^{-1}$ on left and $Q$ on right,
$$\eqalign{
&\quad (-1)^n A^n + c_{n-1} A^{n-1} + \cdots + c_1 A + c_0 I_n\cr
&= (-1)^n Q^{-1}D^nQ + c_{n-1} Q^{-1}D^{n-1}Q + \cdots + c_1 Q^{-1}AQ + c_0 Q^{-1}Q\cr
&= Q^{-1}((-1)^n D^n + c_{n-1} D^{n-1} + \cdots + c_1 D + c_0 I_n)Q\cr}$$
Since $D$ is diagonal, the middle matrix here, 
$$(-1)^n D^n + c_{n-1} D^{n-1} + \cdots + c_1 D + c_0 I_n,$$
is also a diagonal matrix.  The $i$th diagonal entry is
$$(-1)^n \lambda_i^n + c_{n-1} \lambda_i^{n-1} + \cdots + c_1 \lambda_i + c_0.$$
For all $i$, this equals $0$, since it is $\det(A - \lambda_iI) = 0$.
Hence the whole matrix in the middle is the $n\times n$ zero matrix, 
which means the whole product is the zero matrix.
\bigskip
\noindent
IX.  (Eigenvalues in calculus!)
Let $V = C^{\infty}({\bf R})$ be the vector space of all functions
$f : {\bf R} \to {\bf R}$ that have derivatives of all orders at
all $x \in {\bf R}$.
\bigskip
\item{A)} (10)  Let $D : V \to V$ be the linear mapping 
defined by $D(f) = f'$.
Show that for all $\lambda \in {\bf R}$, the eigenspace 
${\rm Ker}(D - \lambda I)$ has dimension 1, and find a basis.
\sol
We have that ${\rm Ker}(D - \lambda I)$ equals the set of 
all functions $f$ satisfying $(D - \lambda I)(f) = 0$.  Here
$I$ means the identity mapping on $V$, so $I(f) = f$ for all $f$.  
We want to find all functions satisfying:
$$D(f) = f' = \lambda f,\ \  {\rm or}\ \  {df\over dx} = \lambda f.$$
This is just the usual exponential growth/decay differential
equation.  We can solve it by separation of variables as
usual:
$$\eqalign{
\int {df \over f}  &= \int \lambda dx\cr
\ln |f| &= \lambda x + k\cr
f &= \pm e^k e^{\lambda x}\cr
  &= ce^{\lambda x},\cr}
$$
where $c \in {\bf R}$ is arbitary.
Hence the eigenspace is one-dimensional,
and $\{e^{\lambda x}\}$ is one basis.
\item{B)} (5)  Let $S : V \to V$ be the linear mapping
defined by $S(f) = f''$.  Let $\lambda = -k^2$ for an
arbitrary $k > 0$.  Find two linearly independent functions in 
the eigenspace ${\rm Ker}(S - \lambda I)$. 
\sol
Recall the derivative formulas
$${d\over dx} \sin(x) = \cos(x); \qquad {d\over dx}\cos(x) = -\sin(x).$$
Thus 
$$S(\sin(x)) = {d^2\over dx^2} \sin(x) = -\sin(x); \qquad S(\cos(x)) = {d^2\over dx^2}\cos(x) = -\cos(x).$$
By the chain rule for derivatives,
$$\{\sin(kx), \cos(kx)\}$$
satisfy 
$$S(\sin(kx)) = {d^2\over dx^2} \sin(kx) = -k^2\sin(kx); \qquad S(\cos(kx)) = {d^2\over dx^2}\cos(kx) = -k^2\cos(kx).$$
So $\{\sin(kx),\cos(kx)\}$ is one such set of functions.
\bigskip
\item{C)} (5)  Let $S : V \to V$ be the linear mapping defined by 
$$S(f) = f(1) + f'(1) (x-1) + {f''(1)\over 2}(x-1)^2.$$  
Show that ${\rm Span}\{1,x,x^2\} \subseteq {\rm Ker}(S - I)$.
For $f(x) = x^2$, 
$$\eqalign{T(f) &= f(1) + f'(1)(x - 1) + {f''(1)\over 2}(x - 1)^2\cr
&= 1 + 2\cdot (x - 1) + {2\over 2}(x - 1)^2\cr
&= 1 + 2x - 2 + x^2 - 2x + 1\cr
&= x^2\cr}$$
Thus $T(f) = f$ and $f$ is an eigenfunction of $T$ with eigenvalue $1$.
The proofs for $f = 1,x$ are similar, and easier. 
Since ${\rm Ker}(S - I)$ is a subspace of $V$, this shows 
${\rm Span}\{1,x,x^2\} \subseteq {\rm Ker}(S - I)$. 
(Note what this says:  the second degree Taylor polynomial of 
a polynomial of degree $\le 2$ at $a = 1$ {\it equals the polynomial itself}.
The same thing is true for all $a$, and all degrees!  This also 
shows that the inclusion is an equality!)
\bye

 
