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\centerline{Mathematics 244 -- Linear Algebra}
\centerline{Answers -- Final Exam}
\centerline{\it May 11, 2004}
\bigskip
\noindent
I.  
\medskip\noindent
A)  Let $p,q\in W$.  Then $p'(1) = q'(1) = 0$, 
$p(0) = 3p(2)$, and $q(0) = 3q(2)$.  So if $c$ is any constant,
then $(cp+q)'(1) = cp'(1)+q'(1) = c\cdot 0 + 0$.  Also,
$$(cp+q)(0) = cp(0) + q(0) = c\cdot 3p(2) + 3q(2) = 3(cp(2) + q(2))
= 3(cp+q)(2) .$$
Since $W$ is closed under sums and scalar multiples, $W$ is a 
subspace of $P_4({\bf R})$.  
\bigskip\noindent
B)  Let $p = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$ be a
 general element of $P_4({\bf R})$.  Then $p \in W$ if and only if
$$\eqalign{
p'(1) &= a_1 + 2a_2 + 3a_3 + 4a_4 = 0\cr
p(0) &= a_0 = 3p(2) = 3a_0 + 6a_1 + 12 a_2 + 24 a_3 + 48 a_4\cr}$$
This says that the coefficients of $p$ must satisfy
$$\matrix{
 &     &  &a_1 &+ &2a_2 &+ &3a_3 &+ &4a_4   &=  &0\cr
 &2a_0 &+ &6a_1 &+ &12 a_2 &+ &24 a_3 &+ &48 a_4 &= &0\cr}$$
 Putting the system into echelon form, we get
$$\matrix{
 & a_0 &     &  &     & + &3a_3 &+ &12a_4&=  &0\cr
 &     & a_1 &+ &2 a_2& + &3a_3 &+ &4a_4 &= &0\cr}$$
The free variables are $a_2, a_3, a_4$, so by the usual process
we get 
$$W = {\rm Span}\{x^2 - 2x, x^3 - 3x - 3 , x^4 - 4x - 12\}$$
(the first comes by setting $a_2 = 1, a_3 = 0, a_4 = 0$, etc.)
\bigskip
\noindent
C)  This set is linearly independent, since if 
$c_1(x^2 - 2x) + c_2(x^3 - 3x - 3) + c_3(x^4 - 4x - 12) = 0,$
then equating coefficients of $x^2$ shows $c_1=0$, 
equating coefficients of $x^3$ shows $c_2=0$, and equating coefficients 
of $x^4$ shows $c_3=0$.
\bigskip
\noindent
II.
\medskip
\noindent
See class notes.  The proof of (1.6.8) in the text follows almost
the same reasoning as this proof.
\bigskip
\noindent
III.  A)  
$$[T]_{std}^{std} = \pmatrix{4&-3&2&-1\cr
                             0&1&1&0\cr
                             1&0&0&-1\cr}$$
since $T(e_1) = (4,0,1) = 4e_1 + 0e_2 + 1e_3$, etc.
\bigskip
\noindent
B)  We have
$$[T]_{\alpha'}^{\beta'} = [I_{{\bf R}^3}]_{std}^{\beta'} 
[T]_{std}^{std}[I_{{\bf R}^4}]^{std}_{\alpha'} $$
From the given bases $\alpha', \beta'$,
$$[I_{{\bf R}^4}]^{std}_{\alpha'} = \pmatrix{1&1&0&0\cr
                                             -1&1&0&0\cr
                                             0&0&2&-1\cr
                                             0&0&3&-2\cr}$$
By the Gauss-Jordan procedure,
$$[I_{{\bf R}^3}]_{std}^{\beta'}
= \pmatrix{0&0&-1\cr
           0&-1&1\cr
           1&1&0\cr}^{-1} = \pmatrix{1&1&1\cr -1&-1&0\cr -1&0&0\cr}$$
So,
$$[T]_{\alpha'}^{\beta'} = 
\pmatrix{1&1&1\cr -1&-1&0\cr -1&0&0\cr}
\pmatrix{4&-3&2&-1\cr 0&1&1&0\cr 1&0&0&-1\cr}
\pmatrix{1&1&0&0\cr -1&1&0&0\cr 0&0&2&-1\cr 0&0&3&-2\cr}$$
$$=\pmatrix{7 & 3 & 0 & -7 \cr
            -6& -2 & -3& 5\cr
            -7&-1&-1&-4\cr}$$
\bigskip
\noindent
IV.  The corresponding echelon form system is
$$\pmatrix{
1 & 0 & {6/7} & {-1/7} &| & {-2/7}\cr
0 & 1 & {3/7} & {3/7} & | & {6/7}\cr
0&0&0&0 & | &a - 10\cr}$$
This means the system is consistent if and only if $a = 10$.
If so the solutions of the system are all vectors
$$\left({-2\over 7},{6\over 7},0,0\right) + s\left({-6\over 7},{-3\over 7},1,0\right) 
+ t\left({1\over 7},{-3\over 7},0,1\right)$$
where $s,t$ are arbitrary real numbers.
\bigskip
\noindent
V.  The base case is when $A = (a_{11})$.  By definition,
$\det(A) = a_{11}$ is the product of the diagonal entries in
this case.  Now, assume that the result is true for all 
$k\times k$ upper-triangular matrices and consider an 
upper-triangular $A$ which is $(k+1) \times (k+1)$.
expanding along the first column, we see
$\det(A) = a_{11} det(A_{11}) + 0$, since all the other
entries in column 1 are zeroes.  The (1,1)-minor $A_{11}$
is a $k\times k$ upper-triangular matrix with diagonal
entries $a_{22}, \ldots, a_{k+1,k+1}$, so 
the induction hypothesis shows that
$$\det(A) = a_{11}\det(A_{11}) = a_{11} a_{22} \cdots a_{k+1,k+1}$$
which establishes the induction step.  Hence the result is true
for all upper-triangular matrices.
\bigskip
\noindent
VI.  A)  Expanding $\det(A - \lambda I)$ along row 3, we have
$$\det\pmatrix{2-\lambda & 9 & 0\cr
               1&2-\lambda&-1\cr
               0&0&5-\lambda\cr} = 
               (5-\lambda)\det\pmatrix{2-\lambda & 9\cr 1&2-\lambda\cr}
               =-(5-\lambda)^2(\lambda+1)$$
so the eigenvalues are $\lambda = 5,5,-1$.
\bigskip
\noindent
B)  We have 
$$\eqalign{
E_5 &= {\rm Ker}\pmatrix{-3 & 9 & 0\cr 1&-3 & -1\cr 0 & 0& 0\cr}\cr
    &= {\rm Ker}\pmatrix{1 & -3 & 0\cr 0&0& 1\cr 0 & 0& 0\cr}\cr
    &= {\rm Span}\{(3,1,0)\}\cr}$$
Similarly,
$$
\eqalign{E_{-1} &= {\rm Ker}\pmatrix{3 & 9 & 0\cr 1&3 & -1\cr 0 & 0& 0\cr}\cr
    &= {\rm Ker}\pmatrix{1 & 3 & 0\cr 0&0& 1\cr 0 & 0& 0\cr}\cr
    &= {\rm Span}\{(-3,1,0)\}\cr}$$
\bigskip
\noindent
C) The multiplicity of $\lambda = 5$ as a root of the characteristic
polynomial is 2, but $\dim E_5 = 1$.  Hence there is no 
way to get a linearly independent set of 3 eigenvectors
of this $A$.  Hence $A$ is not diagonalizable.
\bigskip
\noindent
VII.  A)  True:  Because of the bilinearity of the dot product,
$$\eqalign{
T(cx + w) &= \langle cx + w,y\rangle y + \langle cx + w,z\rangle z\cr
&=c \langle x,y\rangle y +  \langle w,y\rangle y + 
c \langle x,z\rangle z +  \langle w,z\rangle z\cr
&=c(\langle x,y\rangle y + \langle x,z\rangle z) + 
\langle w,y\rangle y + \langle w,z\rangle z\cr
&= cT(x) + T(w)\cr}$$
\bigskip
\noindent
B) True, since the matrix is symmetric (see (4.5.6) in the text).
(Many people worked {\it much, much too hard} on this one!)
\bigskip
\noindent
C)  False.  If $T$ is not surjective, then $\dim {\rm Im}(T) < 2$
and the Dimension Theorem says $\dim {\rm Ker}(T) > 3$.  In
this case, then 
$$\cases{
y \notin {\rm Im}(T) \Rightarrow & no solutions at all\cr
y \in {\rm Im}(T) \Rightarrow & solutions contain more than 3 arbitrary consts.}$$
For example, say $T(x_1,x_2,x_3,x_4,x_5) = (x_1,x_1)$
is a mapping $T : {\bf R}^5 \to {\bf R}^2$.  The image consists
of all vectors whose components are equal.  If $(1,1)$ is such
a vector then 
$T(x_1,x_2,x_3,x_4,x_5) = (1,1)$
has general solution
$$(x_1,x_2,x_3,x_4,x_5) = (1,0,0,0,0) + s(0,1,0,0,0) + t(0,0,1,0,0)
+ u(0,0,0,1,0) + v(0,0,0,0,1)$$
where $s,t,u,v$ are arbitrary.  For other vectors like
$(2,3)$, $T(x_1,x_2,x_3,x_4,x_5) = (2,3)$ has no solutions at all.
\bigskip
\noindent
VIII. A)  The main point of this problem is to derive a way to 
write down the desired polynomials.  The most economical way to do 
this is to note that $p_1$ must have roots at $x = 1,2$,
$p_2$ must have roots at $x = 0,2$,  
etc.  This says
$$p_1(x) = c_1(x-1)(x-2),\ p_2(x) = c_2x(x-2),\ p_3(x) = c_3x(x-1)$$
for some constants $c_i$.  Then the values of $c_i$ can be determined
by substituting the other $x$-value and using the given information
$$p_1(0) = 1 \Rightarrow c_1(0-1)(0-2) = 1 \Rightarrow c_1 = 1/2$$
and similarly, $c_2 = -1$, $c_3 = 1/2$.  The polynomials are
$$\eqalign{
p_1(x) &= {1\over 2}x^2 - {3\over 2}x + 1\cr
p_2(x) &= -x^2 + 2x\cr
p_3(x) &= {1\over 2}x^2 - {1\over 2}x\cr}$$
To check that these are linearly independent, we consider a linear
combination,
$$0 = a_1p_1(x) + a_2 p_2(x) + a_3p_3(x)$$
Substitute $x = 0$, then we get $0 = a_1\cdot 1 + a_2\cdot 0 + a_3\cdot 0$.
Hence $a_1 = 0$.  Similarly, substituting $x = 1$ shows $a_2 = 0$,
$x=2$ shows $a_3 = 0$.  Since $\dim P_2({\bf R}) = 3$, and
this is an independent subset with 3 elements, it is a basis.
\bigskip
B) Write 
$$q(x) =  a_1p_1(x) + a_2 p_2(x) + a_3p_3(x)$$
As above, substitute $x = 0, 1,2$ one after another.  $x = 0$ gives
$$8 = q(0) = a_1\cdot 1 + a_2 \cdot 0 + a_3 \cdot 0$$
Hence $a_1 = 8$.  Similarly, $a_2 = -3$, $a_3 = 4$.
\bigskip
\noindent
C)  Following the idea of proof given in part A above,
$$\eqalign{
p_1(x) = {(x - b)(x - c)\over (a - b)(a - c)}\cr
p_2(x) = {(x - a)(x - c)\over (b - a)(b - c)}\cr
p_1(x) = {(x - a)(x - b)\over (c - a)(c - b)}\cr}$$
This is a special case of a general result called the 
{\it Lagrange Interpolation Theorem}.
\bigskip
\noindent
VIII'.  A)  If $Ax= \lambda x$, then 
$$A^2x = A(Ax) = A(\lambda x) = \lambda Ax = \lambda^2 x$$
Hence $x$ is also an eigenvector of $A^2$ with eigenvalue
$\lambda^2$.
\bigskip
\noindent
B)  Since $A$ is symmetric, so is $A^2$ (reason:  $(A^2)^t = 
(AA)^t = A^t A^t = A^2$).  Apply the Spectral Theorem to the
symmetric matrix $A^2$.
Part A says that the eigenspaces
and the projections for $A^2$ are the same as those for $A$
(the eigenspaces are equal).  But the eigenvalues for $A^2$ 
are the $\lambda^2$.  Hence the spectral decomposition for $A^2$
is 
$$A^2 = \lambda_1^2 P_1 + \cdots + \lambda_k^2 P_k$$
(Alternate method:  From the spectral decomposition for
$A$, 
$$A = \lambda_1 P_1 + \cdots + \lambda_k P_k$$
So 
$$A^2 = (\lambda_1 P_1 + \cdots + \lambda_k P_k)(\lambda_1 P_1 + 
\cdots + \lambda_k P_k)$$
Matrix product is distributive over matrix sum, so this can be
written as 
$$A^2 =\lambda_1^2 P_1^2 + \cdots + \lambda_k^2 P_k^2 + \sum_{i\ne j}\lambda_i\lambda_j P_iP_j$$
But $P_i^2 = P_i$ and if $i \ne j$, then $P_iP_j = 0$.  This follows because
as we know, eigenvectors for distinct eigenvalues of a symmetric 
matrix are orthogonal, hence
${\rm Im}(P_j) = E_{\lambda_j} \subset (E_{\lambda_i})^\perp = 
{\rm Ker}(P_i)$.  So when you apply $P_iP_j$ to any vector you 
will get zero.)
\bigskip
\noindent
C)  The eigenvalues of $A$ are $\lambda = 3,5$.  Correxponding
bases for the eigenspaces are $\{(1,1)\}$ for $E_5$
and $\{(-1,1)\}$ for $E_3$.
$$P_{E_5} = \pmatrix{1\cr 1\cr}(\pmatrix{1 &1\cr}\pmatrix{1\cr 1\cr})^{-1} \pmatrix{1 & 1\cr}
= \pmatrix{1/2 & 1/2 \cr 1/2 & 1/2 \cr}$$
Similarly,
$$P_{E_3} = \pmatrix{-1\cr 1\cr}(\pmatrix{-1 &1\cr}\pmatrix{-1\cr 1\cr})^{-1} \pmatrix{-1 & 1\cr}
= \pmatrix{1/2 & -1/2 \cr -1/2 & 1/2 \cr}$$
So 
$$A^2 = \pmatrix{17 & 8 \cr 8& 17\cr} = 25 \pmatrix{1/2 & 1/2 \cr 1/2 & 1/2 \cr}
+ 9\pmatrix{1/2 & -1/2 \cr -1/2 & 1/2 \cr}$$
\bye