\magnification=\magstep1
\def\sol{\bigskip\item{}{\it Solution:\ }}
\centerline{Mathematics 244, section 1 -- Linear Algebra}
\centerline{Exam 3 Solutions -- April 30, 2007}
\bigskip
\noindent
I.  Let $V = {\rm Span}\{e^x,\cos(x),\sin(x)\}$,
let $W = P_2({\bf R})$ and 
let $T : V\to W$ be the mapping defined by 
$$T(f) = f(0) + f'(0)x + {f''(0)\over 2}x^2.$$
\bigskip
\item{A)} (10)  Show that $T$ is a linear mapping.
\sol
Let $f,g \in V$.  Then 
$$\eqalign{T(f+g) &=  (f+g)(0) + (f+g)'(0)x + {(f+g)''(0)\over 2}x^2\cr
        &=  f(0) + f'(0)x + {f''(0)\over 2}x^2 + g(0) + g'(0)x + {g''(0)\over 2}x^2\cr
        &= T(f) + T(g).\cr}$$
Similarly, for $c \in {\bf R}$,
$$\eqalign{T(cf) &=  (cf)(0) + (cf)'(0)x + {(cf)''(0)\over 2}x^2\cr
        &=  cf(0) + cf'(0)x + c{f''(0)\over 2}x^2\cr
        &= cT(f).\cr}$$
\item{B)} (15) Let $B = \{e^x,\cos(x),\sin(x)\}$ (a basis
for $V$) and $C = \{1,x,x^2\}$ (basis for $W$).  Find the matrix
$[T]_{C \leftarrow B}$.
\sol
Note that $T(f)$ is just the second-degree Taylor polynomial
of $f$ at $a = 0$.  So we can compute $T(f)$ using facts from
calculus:
$$\eqalign{T(e^x) &= 1 + x + {x^2\over 2} = 1\cdot 1 + 1\cdot x + {1\over 2}\cdot x^2\cr
T(\cos(x)) &= 1 - {x^2\over 2} = 1\cdot 1 + 0\cdot x + {-1\over 2} \cdot x^2\cr
T(\sin(x)) &= x = 0\cdot 1 + 1\cdot x + 0\cdot x^2\cr}$$
Hence placing the coordinate vectors of these $T(f)$ into the columns,
$$[T]_{C \leftarrow B} = \pmatrix{1 & 1 & 0\cr
                                  1 & 0 & 1\cr
                                  {1\over 2} & {-1\over 2} & 0\cr}.$$ 
\item{C)} (15)  Now, let $C' = \{x^2 + x, x - 2, x + 1\}$.  
Find the matrix $[T]_{C' \leftarrow B}$.
\sol 
We have 
$$[T]_{C'\leftarrow B} = P_{C' \leftarrow C} [T]_{C \leftarrow B}.$$
The matrix 
$$P_{C' \leftarrow C} = \left(P_{C \leftarrow C'}\right)^{-1} = 
\pmatrix{0 & -2 & 1\cr
         1 &  1 & 1\cr
         1 &  0 & 0\cr}^{-1}.$$
Computing the inverse,
$$\eqalign{
\pmatrix{0 & -2 & 1 &|& 1 & 0 & 0\cr
         1 &  1 & 1 &|& 0 & 1 & 0\cr
         1 &  0 & 0 &|& 0 & 0 & 1\cr} 
&\sim \pmatrix{1 &  0 & 0 &|& 0 & 0 & 1\cr
               0 & -2 & 1 &|& 1 & 0 & 0\cr
               1 &  1 & 1 &|& 0 & 1 & 0\cr} \cr
&\sim \pmatrix{1 &  0 & 0 &|& 0 & 0 & 1\cr
               0 & -2 & 1 &|& 1 & 0 & 0\cr
               0 &  1 & 1 &|& 0 & 1 & -1\cr} \cr
&\sim \pmatrix{1 &  0 & 0 &|& 0 & 0 & 1\cr
               0 &  1 & -1/2&|& -1/2 & 0 & 0\cr
               0 &  1 &   1  &|& 0   & 1 & -1\cr} \cr
&\sim \pmatrix{1 &  0 & 0 &|& 0 & 0 & 1\cr
               0 &  1 & -1/2 &|& -1/2 & 0 & 0\cr
               0 &  0 &   3/2  &|& 1/2 & 1 & -1\cr} \cr
&\sim \pmatrix{1 &  0 & 0 &|& 0 & 0 & 1\cr
               0 &  1 &  0 &|& -1/3 & 1/3 & -1/3\cr
               0 &  0 &  1  &|& 1/3 & 2/3 & -2/3\cr} \cr}$$
So
$$\pmatrix{0 & -2 & 1\cr
           1 &  1 & 1\cr
           1 &  0 & 0\cr}^{-1} = \pmatrix{0 & 0 & 1\cr
                                        -1/3 & 1/3 & -1/3\cr
                                        1/3 & 2/3 & -2/3\cr}.$$
Then 
$$[T]_{C'\leftarrow B} = 
\pmatrix{0 & 0 & 1\cr
    -1/3 & 1/3 & -1/3\cr
     1/3 & 2/3 & -2/3\cr} \pmatrix{1 & 1 & 0\cr
                                   1 & 0 & 1\cr
                               1/2 & -1/2 & 0\cr} = \pmatrix{1/2 & -1/2 & 0\cr
                                                             -1/6 & -1/6 & 1/3\cr
                                                              2/3 & 2/3 & 2/3\cr}.$$
\bigskip
\noindent
II.  Let 
$$A = \pmatrix{1   & s &  s\cr
               -1   & s &  s\cr
               -1   & -1 &  1\cr}$$
The characteristic
polynomial of $A$ is 
$$p(\lambda) = -\lambda^3 + (s + 2)\lambda^2 - (5s + 1)\lambda + 4s.$$
\item{A)} (10) Determine a value of $s$ such
that $\lambda = 3$ will be an eigenvalue of $A$.
\sol
This will be the case when
$$0 = -27 + 9(s+2) - 3(5s + 1) + 4s = -2s - 12,$$
so $s = -6$.  
\bigskip
\item{B)}  (20)  Let $s = -1$.  Is $A$ diagonalizable in 
this case?  If so, find an invertible matrix $Q$ and a diagonal
matrix $D$ such that $Q^{-1}AQ = D$.  If not, say why not.
{\it Note:}  You may ``purchase'' a factored
form of the characteristic polynomial with $s = -1$
from me during the exam for a ``price'' of 5 points on this part. 
\sol
When $s = -1$, the characteristic polynomial is 
$$-\lambda^3 + \lambda^2 + 4\lambda - 4 = (\lambda - 1)(4 - \lambda^2) = 
(\lambda - 1)(2 - \lambda)(2 + \lambda).$$
The eigenvalues are $\lambda = 1, 2, -2$.  These are distinct, so 
we know $A$ is diagonalizable.  (Note this also follows from the 
Spectral Theorem, since $A$ is symmetric when $s = -1$(!))
We have for $\lambda = 1$,
$${\rm Nul}(A - I) = {\rm Nul}\pmatrix{0 & -1 & -1\cr
                                 -1 & 2 & -1\cr
                                 -1 & -1 & 0\cr} = {\rm Span}\left\{\pmatrix{1 \cr 1\cr 1\cr}\right\}.$$
For $\lambda = 2$,
$${\rm Nul}(A - 2I) = {\rm Nul}\pmatrix{-1 & -1 & -1\cr
                                 -1 & 3 & -1\cr
                                 -1 & -1 & -1\cr} = 
{\rm Span}\left\{\pmatrix{-1 \cr 0\cr 1\cr}\right\}.$$
Finally, for $\lambda = -2$,
$${\rm Nul}(A + 2I) = {\rm Nul}\pmatrix{3 & -1 & -1\cr
                                        -1 & 1 & -1\cr
                                        -1 & -1 & 3\cr} = 
{\rm Span}\left\{\pmatrix{1 \cr 2\cr 1\cr}\right\}.$$
So 
$$D = \pmatrix{1 & 0 & 0\cr
               0 & 2 & 0\cr
               0 & 0 & -2\cr} \quad {\rm and} \quad 
Q = \pmatrix{1 & -1 & 1\cr
             1 &  0 & 2\cr 
             1 &  1 & 1\cr}.$$
 
\bigskip
\noindent
III.  (True/False)  For each true statement, give a short proof.
For each false statement, give a counterexample.  Do any {\it three} parts.
If you submit solutions for all four, I will count any additional points you 
earn as extra credit.  
\bigskip
\item{A)} (10) Let $A$ be an $n\times n$ matrix.
If $\{v_1,\ldots,v_k\}$ are eigenvectors with
distinct eigenvalues $\lambda_i$, $i = 1, \ldots, k$,
then $\{v_1,\ldots,v_k\}$ is an orthogonal set.
\sol
This is {\it False}.  A counterexample:  Let $A = \pmatrix{1 & 1\cr 0 & 2}$.
Then $\lambda = 1, 2$ are eigenvalues, with corresponding
eigenvectors $v_1 = \pmatrix{1 \cr 0\cr}$ and $v_2 = \pmatrix{1 \cr 1\cr}$.
But $v_1 \cdot v_2 = 1 \ne 0$.  (The statement is true if $A$ is symmetric, though.)
\bigskip 
\item{B)} (10)  If $A$ and $B$ are $n\times n$ matrices with
the same characteristic polynomial, then $A$ and $B$ are similar.
\sol
This is {\it False}.  A counterexample:  Let $A = \pmatrix{1 & 1\cr 0 & 1\cr}$
and $B = \pmatrix{1 & 0 \cr 0 & 1\cr}$.  Both $A$ and $B$ have
characteristic polynomial $(\lambda - 1)^2$.  But $A$ is not diagonalizable
while $B$ is diagonalizable (diagonal, even).  Hence they cannot be similar.
(The converse of this statement is true.)
\bigskip
\item{C)}  (10) If $v$ is a nonzero vector in ${\bf R}^n$, and
$P_v$ is the orthogonal projection mapping to the line spanned by 
$v$, then the matrix $[P_v]_{{\cal B} \leftarrow {\cal B}}$ 
has rank 1 no matter which basis ${\cal B}$ is used in ${\bf R}^n$.
\sol
This is {\it True}.  $P_v(w) = {v\cdot w\over v\cdot v} v$ is a
multiple of $v$ for all $w$.  Hence if ${\cal B} = \{b_1, \ldots, b_n\}$
is any basis, then for all $i$, $P_v(b_i)$ are scalar multiples of $v$.
Hence $[P_v(b_i)]_{\cal B}$ 
will all be scalar multiples of $[v]_{\cal B}$.
It follows that the dimension of the column space of 
$[P_v]_{{\cal B} \leftarrow {\cal B}}$
is $1$ for all bases ${\cal B}$. 
\bigskip
\item{D)}  (10)  If a $6\times 6$ matrix $A$ has 
characteristic polynomial $(1 - \lambda)^3(6 - \lambda)^3$, and
${\rm rank}(A - I) = 4$ and ${\rm rank}(A - 6I) = 3$, then
$A$ is diagonalizable.  
\sol
This is {\it False}.  Since ${\rm rank}(A - I) = 4$, $\dim {\rm Nul}(A - I) = 2$.
Similarly, since ${\rm rank}(A - 6I) = 3$, $\dim {\rm Nul}(A - 6I) = 3$.
Since $2 + 3 < 6$, $A$ cannot be diagonalizable.
\bye
