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\noindent
{\Large Math 211 \qquad Homework 13 \hfill Sample Texfile}


\medskip\centerline{\bf 10 points for each question.}


\medskip
\noindent{\bf 1.}
Let 
$${\bf y}={\small \left[\matrix{-1\cr 4\cr 3}\right]} \,\,\, , \,\,\,
{\bf u}_1={\small \left[\matrix{1\cr 1\cr 1}\right]} \,\,\, , \,\,\,
{\bf u}_2={\small \left[\matrix{-1\cr 3\cr -2}\right]}.
$$

(a) Show that $\{{\bf u}_1, {\bf u}_2\}$ is
an orthogonal set.

(b) Express ${\bf y} = \hat {\bf y} + {\bf z}$ so that $\hat {\bf y}$ is
orthogonal projection of the vector ${\bf y}$ onto 
$W = $Span$\{{\bf u}_1, 
{\bf u}_2\}$.

(c) Find the distance of ${\bf y}$ to the subspace $W$.

(d) Normalize the vectors in $\{{\bf u}_1, {\bf u}_2, {\bf z}\}$ to get orthonormal 
basis.

\medskip
\noindent{\bf 2.}
Let 
$${\bf u}_1 = \left[\matrix{3\cr 4\cr 0\cr 1\cr}\right]
\quad \hbox{ and } 
{\bf u}_2=  \left[\matrix{-4\cr 3\cr 1\cr 0\cr}\right].$$

(a) Show that $\{{\bf u}_1, {\bf u}_2\}$ is
an orthogonal set.

(b) Express ${\bf y} = \hat {\bf y} + {\bf z}$ so that $\hat {\bf y}$ is
orthogonal projection of the vector ${\bf y}=\left[
\matrix{6\cr 3\cr -2\cr 2\cr}\right]$ onto 

\qquad
$W = {\rm Span}\{{\bf u}_1, 
{\bf u}_2\}$.

(c) Find the distance of ${\bf y}$ to the subspace $W$.

(d) Find a nonzero vector ${\bf z}_2$ so that 
$S = \{{\bf u}_1, {\bf u}_2, {\bf z}, {\bf z}_2\}$
is an orthogonal basis for $\IR^4$.

(e) Normalize the vectors in $S$ to get an orthonormal basis.




\medskip
\noindent{\bf 3.}
Let 
$$\mathcal{B}={\small \left\{ \left[\matrix{3\cr -4\cr 5}\right],
\left[\matrix{-3\cr 14\cr -7}\right], \left[\matrix{0\cr 1\cr 0}\right]  \right\}}.$$

(a) Show that ${\mathcal B}$ is a basis for $\IR^3$.

(b) Apply the Gram-Schmidt process to ${\mathcal B}$ to get an orthogonal basis
${\mathcal B}_1$.

(c) Normalize the vectors in ${\mathcal B}_1$ to get and orthnormal basis
${\mathcal B}_2$.

(d) Let $A$ be the matrix with the vectors in ${\mathcal B}$ as columns, and 
let $V$ be the matrix with the 

\quad vectors  in ${\mathcal B}_2$ as columns.
Show that $V^TA = R$ is in upper triangular from.



\medskip
\noindent{\bf 4.}
Let 
$$A={\small \left[\matrix{1&3&5\cr -1&-3&1\cr 0&2&3\cr 1&5&2\cr 1&5&8}
\right]}.$$

(a) Show that $A$ has linearly independent columns.

(b) Apply the  Gram-Schmidt process to the columns of $A$ to get an
orthogonal basis ${\mathcal B}$ for 

\quad the column space of $A$.

(c) Normalize the vectors in ${\mathcal B}$ to get an 
orthonormal basis ${\mathcal U}$ for the column basis for $A$.

(d) Let $Q$ be the matrix with the vectors in ${\mathcal U}$ as columns.
Show that $Q^TA = R$ is in upper 

\quad triangular form.

\quad 
{\bf Remark: We have $A = QR$, which is known as the $QR$ factorization of $A$.}


\medskip
\noindent{\bf 5.} Repeat the procedures in Question 4 for the matrix
$A={\small \left[\matrix{-2&3\cr 5&7\cr 2&-2\cr 4&6}\right]}$
to find the $QR$ factorization 

of $A$.


\end{document}
