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\newcommand{\docdate}{October 3, 2001}
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\newcommand{\professor}{Professors Cholak and Polini}
\newcommand{\coursenum}{Math 221}
\newcommand{\courseblurb}{Linear Algebra}

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\begin{document}

\doctitle


\begin{problem}
  (This is based on problem 76 on page 90.)  The color of light can be
  represented in a vector, $
  \begin{bmatrix}
    R \\ G \\ B \end{bmatrix}$, where $R$ is the amount of red light,
  $G$ is the amount of green light and $B$ is the amount of blue
  light.  The human eye and the brain transforms the incoming signal
  into the signal $\begin{bmatrix} I \\ L \\ S \end{bmatrix}$, where
  the intensity $I = \frac{R+G+B}{3}$, the long-wave signal $L = R - G
  $ and the short wave signal $S =B - \frac{R+G}{2}$.  

  \begin{subproblem}
    Find the matrix $\mathcal{A}$ of the transformation taking $
  \begin{bmatrix}
    R \\ G \\ B \end{bmatrix}$ to $\begin{bmatrix} I \\ L \\ S
  \end{bmatrix}$.
  \end{subproblem}

$\mathcal{A} = 
\begin{bmatrix}
  \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
   1 & -1 & 0 \\
  -\frac{1}{2} & -\frac{1}{2} & 1
\end{bmatrix}$.

\begin{subproblem}
  Find $\mathcal{A}^{-1}$.
\end{subproblem}

$ \mathcal{A}^{-1} = \begin{bmatrix} 1 & \frac{1}{2} & - \frac{1}{3}
  \\ 1 & -\frac{1}{2} & - \frac13 \\ 1 & 0 & \frac23
\end{bmatrix}$.

\begin{subproblem}
  Consider a pair of yellow sunglasses for water sports which cuts out
  all blue light but passes all red and green light.  Find a $3 \times
  3$ matrix $\mathcal{B}$ that represents the transformation incoming
  light undergoes as it passes through the sunglasses.
\end{subproblem}

$\mathcal{B} = 
\begin{bmatrix}
  1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 
\end{bmatrix}$.

\begin{subproblem}
  Is $\mathcal{B}$ invertible?
\end{subproblem}

No.  The rank of $\mathcal{B}$ is 2 which is less than 3.

\begin{subproblem}
  Find the matrix for the composite transformation that light
  undergoes as it first passes through the sunglasses and then the
  eye.
\end{subproblem}

This is $\mathcal{A} \mathcal{B} = \begin{bmatrix}
  \frac{1}{3} & \frac{1}{3} & 0 \\
   1 & -1 & 0 \\
  -\frac{1}{2} & -\frac{1}{2} & 0
\end{bmatrix}$.

\begin{subproblem}
  As you put on the sunglasses, the signal you receive (the $I,L,$ and
  $S$) undergoes a transformation.  Find the matrix $\mathcal{M}$ of
  this transformation.  (There is nice picture in the book if this is
  not clear enough.)
\end{subproblem}

Lets think before begin (always a good thing). We want two
compositions to be the same.  In the first transformation the light
goes though the sunglasses, represented by the matrix $\mathcal{B}$,
and then though the eye represented.  In the second transformation,
light through the eye, represented by $\mathcal{A}$, and then through
an unknown transformation represented by $\mathcal{M}$.  So for all
$\vec{x}$, we want $\mathcal{A}\mathcal{B}\vec{x} =
\mathcal{M}\mathcal{A} \vec{x}$.  Hence we want $\mathcal{M}$ such
that $\mathcal{A}\mathcal{B} = \mathcal{M}\mathcal{A}$.  Since
$\mathcal{A}$ is invertible we can multiple the right side by
$\mathcal{A}^{-1}$ and hence $\mathcal{M} = \mathcal{ABA}^{-1}$.  So
we have the $\mathcal{M}$ is $\begin{bmatrix} \frac12 & 0 & -\frac29
  \\ 0 & 1& 0 \\ -1 & 0 & \frac13
\end{bmatrix}$.


\end{problem}

\begin{problem}
  Let $B$ be a $m \times n$ matrix with $m \geq n$. Let $A$ be a
  matrix such that $BA$ is invertible.  (Hint: look at problems 31-35
  from section 2.4)

\begin{subproblem}
What is the size of $A$?
\end{subproblem}

$BA$ must be square and hence must be $n \times n$.  So $A$ is $n
\times m$.

\begin{subproblem}
  Is $T(\vec{x})= B\vec{x}$ onto? Or equivalently, for all $\vec{b}
  \in {\Bbb R}^m$ there exists a vector $\vec{x} \in {\Bbb R}^n$ such
  that $B\vec{x}= \vec{b}$. Or equivalently, the linear system
  $B\vec{x}= \vec{b}$ is consistent for all vectors $\vec{b} \in {\Bbb
    R}^m$.
\end{subproblem}

Yes, for each vector  $\vec{b} \in {\Bbb R}^m$, take $\vec{x}=A \vec{b}$.

\begin{subproblem}
What is the rank of $B$?
\end{subproblem}

Since the linear system $B\vec{x}= \vec{b}$ is consistent for all vectors 
$\vec{b} \in {\Bbb R}^m$ the rank of $B$ must be $m$. 

\begin{subproblem}
 What is the realtion between $n$ and $m$?
\end{subproblem}

$n=m$ because $m=\text{rank}(B)\leq n \leq m$.

\begin{subproblem}
Apply a lemma from the book (from class) to show that $B$ and 
$A$ are invertible. 
\end{subproblem}

$A$ and $B$ are square matrices. Since the rank of $B$ is $m=n$,
$rref(B)=I_n$. Thus $B$ is invertible. Now multiplying $BA=C$ by
the inverse of $B$, we get $A=B^{-1}C$. Since $B^{-1}$ and $C$
are invertible $A$ is invertible.


\end{problem}



%\newpage
\begin{problem}


\begin{subproblem}
Find the matrix of the linear transformation $T_{\vec{a}} :
{\Bbb R}^3 \rightarrow {\Bbb R}^3$ that projects each vector
$\vec{x}$ onto the line given by 
$\vec{a} = \begin{bmatrix}
  1 \\
  0 \\
  1 \\
\end{bmatrix}$
\end{subproblem}

Call ${\mathcal A}$ the matrix of the linear transformation
$T_{\vec{a}}$. I first need to find the unit vector on the line given
by $\vec{a}$: $\vec{u}= \begin{bmatrix}
  1/\sqrt{2} \\
  0 \\
  1/\sqrt{2} \\
\end{bmatrix}$.
$\mathcal A =
\begin{bmatrix}
   1/2 & 0 &  1/2 \\
   0 & 0 & 0 \\
   1/2 & 0 & 1/2
\end{bmatrix}$.

\begin{subproblem}
Find the matrix of the linear transformation $T_{\vec{b}} :
{\Bbb R}^3 \rightarrow {\Bbb R}^3$ that projects each vector
$\vec{x}$ onto the line given by 
$\vec{b} = \begin{bmatrix}
  1 \\
  -1 \\
  0 \\
\end{bmatrix}$
\end{subproblem}


Call ${\mathcal B}$ the matrix of the  linear transformation
$T_{\vec{b}}$.
I first need to find the unit vector on the line given by
$\vec{b}$: $\vec{v}=  \begin{bmatrix}
  1/\sqrt{2} \\
-1/\sqrt{2}\\
 0 \\
\end{bmatrix}$.
$\mathcal B =
\begin{bmatrix}
   1/2 & -1/2 &  0 \\
   -1/2 & 1/2 & 0 \\
   0 & 0 & 0
\end{bmatrix}$.\newpage

\begin{subproblem}
Now use the above informations to find the matrix of the linear
transformation $T :
{\Bbb R}^3 \rightarrow {\Bbb R}^3$ that projects each vector
$\vec{x}$ onto the plane ${\mathcal P}$ given by $\vec{a}$ and
$\vec{b}$. (The plane ${\mathcal P}$  is the set of all linear
combinations of $\vec{a}$ and
$\vec{b}$.)  (Hint: look at problem 40 from section 2.2)
\end{subproblem}

Call ${\mathcal C}$ the matrix of the linear transformation $T : {\Bbb
  R}^3 \rightarrow {\Bbb R}^3$ that projects each vector $\vec{x}$
onto the plane ${\mathcal P}$ given by $\vec{a}$ and $\vec{b}$.
$T(\vec{x}) = T_a(\vec{x}) + T_b(\vec{x}) = \mathcal{A}\vec{x} +
\mathcal{B}\vec{x}$.  So $\mathcal C = {\mathcal A} + {\mathcal B} =
\begin{bmatrix}
   1 & -1/2 & 1/2  \\
   -1/2 & 1/2 & 0 \\
   1/2 & 0 & 1/2
\end{bmatrix}$. 
%\medskip

\noindent

\end{problem}

\end{document}

