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\newcommand{\docid}{Quiz 6} \renewcommand{\today}{December 1, 2000}

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\begin{document}
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\begin{problem}
  Let $V$ be a subspace of $\mathbf{R}^4$ whose basis is $\vec{x_1}=
  (1,0,1,0,0)$, $\vec{x_2} = (0,1,-1,-1,0)$ and $\vec{x_3} =
  (0,0,10,0,10)$.
  \begin{subproblem}
    Find an orthogonal basis $B$ for $V$. (Here it might be useful to
    multiply one of the vectors you get to avoid fractional entries.)
  \end{subproblem}
  \begin{subproblem}
    Find an orthonormal basis for $V$.
  \end{subproblem}
  \begin{subproblem}
    Find a basis for $V^\perp$.  (Hint: The orthogonal complement of
    the rowspace of a matrix $A$ is what?)
  \end{subproblem}
  \begin{subproblem}
    Find an orthogonal basis for $V^\perp$.
  \end{subproblem}
  \begin{subproblem}
    Expand the basis $B$ found in the first part to an orthogonal
    basis of $\mathbb{R}^5$. (This does not involve any computations.)
  \end{subproblem}
  \begin{subproblem}
    Let $\vec{x} = (1,2,1,0,1)$. Find $\text{proj}_V(\vec{x})$,
    $\vec{z} = \vec{x} - \text{proj}_V(\vec{x})$ and
    $[\text{proj}_V(\vec{x})]_B$.
  \end{subproblem}
  (Expect for part (b) all the numbers work out nicely.)
\end{problem}

%(a) Let $\vec{v}_1 = \vec{x}_1$. Then $\vec{v}_2 = \vec{x}_2 -
%\text{proj}_{W_1}(\vec{x_2}) = \vec{x}_2 - \frac{\vec{v}_1 \cdot
%  \vec{x}_2} {\vec{v}_1 \cdot \vec{v}_1} \vec{v}_1$, where $W_1 =
%\text{Span}\{ \vec{v}_1\}$.  So $\vec{v}_2 = (0,1,-1,-1,0) -
%\frac{-1}{2}(1,0,1,0,0) = (\frac{1}{2},1,-\frac{1}{2},-1,0)$.  To make
%life easier lets multiply $\vec{v}_2$ by $2$ to get $\vec{v}_2 =
%(1,2,-1,-2,0)$.  Then $\vec{v}_3 = \vec{x}_3 -
%\text{proj}_{W_2}(\vec{x_3}) = \vec{x}_3 - \frac{\vec{v}_1 \cdot
%  \vec{x}_3} {\vec{v}_1 \cdot \vec{v}_1} \vec{v}_1 - \frac{\vec{v}_2
%  \cdot \vec{x}_3} {\vec{v}_2 \cdot \vec{v}_2} \vec{v}_2$, where $W_2
%= \text{Span}\{ \vec{v}_1, \vec{v}_1\}$.  Hence $\vec{v}_3 =
%(0,0,10,0,10) - \frac{10}{2}(1,0,1,0,0) - \frac{-10}{10}(1,2,-1,-2,0)
%= (0,0,10,0,10) - (5,0,5,0,0) + (1,2,-1,-2,0) = (-4,2,4,-2,10)$.
%Hence an orthogonal basis is for $V$ is $ (1,0,1,0,0)$,
%$(1,2,-1,-2,0)$, $(-4,2,4,-2,10)$.

%(b) Just normalize the above basis to get
%$(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},0,0)$,
%$\frac{1}{\sqrt{10}}(1,2,-1,-2,0)$,
%$\frac{1}{\sqrt{140}}(-4,2,4,-2,10)$.

%(c) Let $A$ be the matrix whose rows are the above vectors, $A =
%\begin{bmatrix}
%  1 & 0 & 1 & 0 & 0\\
%  0 & 1 & -1 & -1 & 0 \\
%  0 & 0 & 10 & 0 & 10 \\
%\end{bmatrix}$.  So $V$ is the row space of $A$ and $V^\perp$ is the
%nullspace of $A$.  The reduced row echelon form of $A$ is
%$\begin{bmatrix}
%  1 & 0 & 0 & 0 & 1 \\
%  0 & 1 & 0 & -1 & 1 \\
%  0 & 0 & 1 & 0 & -1 \\
%\end{bmatrix}$.  
%So a basis for $V^\perp$ are the two vectors $\vec{x}_4 =(0,1,0,1,0)$
%and $\vec{x}_5 = (1,-1,-1,0,1)$.

%(d).  We will use $\vec{v}_4$ and $\vec{v}_5$ rather than $\vec{v}_1$
%and $\vec{v}_2$.  Let $\vec{v}_4 = \vec{x}_4 = (0,1,0,1,0)$. Then
%$\vec{v}_5 = \vec{x}_5 - \text{proj}_{W_4}(\vec{x_5}) = \vec{x}_5 -
%\frac{\vec{v}_4 \cdot \vec{x}_5} {\vec{v}_4 \cdot \vec{v}_4}
%\vec{v}_4$, where $W_4 = \text{Span}\{ \vec{v}_4\}$.  So $\vec{v}_2 =
%(1,-1,-1,0,1) - \frac{-1}{2}(0,1,0,1,0) =
%(1,-\frac{1}{2},-1,\frac{1}{2},1)$.  Again to make life easier lets
%multiply $\vec{v}_2$ by $2$ to get $\vec{v}_2 = (2,-1,-2,1,2)$.

%(e) Since the element of the basis in part (d) is orthogonal to the
%elements in the basis in part (a).  The union of the two basis is the
%desired basis. $ (1,0,1,0,0)$, $(1,2,-1,-2,0)$, $(-4,2,4,-2,10)$,
%$(0,1,0,1,0)$, $(2,-1,-2,1,2)$.

%(f) $\text{proj}_V(\vec{x}) = \frac{\vec{v}_1 \cdot \vec{x}}{\vec{v}_1
%  \cdot \vec{v}_1}\vec{v}_1 + \frac{\vec{v}_2 1\cdot
%  \vec{x}}{\vec{v}_2 \cdot \vec{v}_2}\vec{v}_2 + \frac{\vec{v}_3
%  1\cdot \vec{x}}{\vec{v}_3 \cdot \vec{v}_3}\vec{v}_3 = \frac{2}{2}
%\vec{v}_1 + \frac{4}{10}\vec{v}_2 + \frac{14}{140} \vec{v}_3 =
%(1,1,1,-1,1)$. Hence $[\text{proj}_V(\vec{x})]_B =
%(1,\frac{1}{5},\frac{1}{10})$. $\vec{z} = (1,2,1,0,1) - (1,1,1,-1,1) =
%(0,1,0,1,0)$.



\end{document}

