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\centerline{Mathematics 244, section 1 -- Linear Algebra}
\centerline{Exam 3 -- April 27, 2007}
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\noindent
{\it Directions}
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Do all work in the blue exam booklet -- there are 100 possible points,
and 10 possible Extra Credit points (see question III).
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\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 
$$T(f) = f(0) + f'(0)x + {f''(0)\over 2}x^2.$$
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\item{A)} (10)  Show that $T$ is a linear mapping.
\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
$[T]_{C \leftarrow B}$.
\item{C)} (15)  Now, let $C' = \{x^2 + x, x - 2, x + 1\}$.  
Find the matrix $[T]_{C' \leftarrow B}$.
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\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 $A$.
\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.  
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\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.  
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\item{A)} (10) Let $A$ be an $n\times n$ matrix.
If $\{v_1,\ldots,v_k\}$ are eigenvectors of $A$ with
distinct eigenvalues $\lambda_i$, $i = 1, \ldots, k$,
then $\{v_1,\ldots,v_k\}$ is an orthogonal set.
\item{B)} (10)  If $A$ and $B$ are $n\times n$ matrices with
the same characteristic polynomial, then $A$ and $B$ are similar.
\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 ${\bf R}^n$.
\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.  
\bye
