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MATH69.217* ANSWERS TO TEST 3 Fall 2001


1.
(a) State the definition of an orthonormal basis in ${\bf R}^n$. (b) Verify the identity

\begin{displaymath}{\bf v}\bullet {\bf w}={1\over 4}\, \left(
\Vert{\bf v}+{\bf w}\Vert^2- \Vert{\bf v}-{\bf w}\Vert^2\right)\, . \end{displaymath}

where ${\bf u}\bullet {\bf v}$ stands for the dot product of vectors $\bf u$ and $\bf w$ in ${\bf R}^n$.


Answer:
(a) An orthonormal basis in ${\bf R}^n$ is a basis in ${\bf R}^n$ consisting of unit vectors perpendicular to each other. (b) We begin with the right hand side:

\begin{eqnarray*}
& & {1\over 4}\, \left(\Vert{\bf v}+{\bf w}\Vert^2-\Vert{\bf ...
...{1\over 4}\ \, 4\,{\bf v}\bullet {\bf w}={\bf v}\bullet {\bf w}
\end{eqnarray*}



Done.


2.
Find the entries $x$, $y$ and $z$ in the following orthogonal matrix

\begin{displaymath}A={1\over 5}\, \left[\begin{array}{ccc}
4 & 3/\sqrt{2} & 3/\...
...{2} \\
3 & x & y \\
0 & 5/\sqrt{2} & z \end{array}\right]. \end{displaymath}


Answer:
From the fact that the first two columns are perpendicular, we obtain $x=-4/\sqrt{2}=-2\sqrt{2}$. From the fact that the first and the last rows are perpendicular, we get $z=-5/\sqrt{2}$. Finally, from the fact last two rows are perpendicular, we get $y=-2\sqrt{2}$.


3.
Find the projection of the vector ${\bf v}$ onto the subspace spanned by the vectors ${\bf w}_1$ and ${\bf w}_2$, where

\begin{displaymath}{\bf v}=\left[\begin{array}{r} 3 \\ 2 \\ 2 \end{array}\right]...
...f w}_2=\left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right]. \end{displaymath}

Begin by checking that $\{{\bf w}_1,\, {\bf w}_2\}$ is an orthogonal system.


Answer:
First we check ${\bf w}_1\bullet{\bf w}_2=1\times 1 +1\times 1 +2\times (-1)=0$. Hence $\{{\bf w}_1, \ {\bf w}_2\}$ is indeed and orthogonal system. The required projection is

\begin{displaymath}
\frac{{\bf v}\bullet{\bf w}_1}{{\bf v}\bullet{\bf w}_1}\ {\...
...
=\left[\begin{array}{r} 5/2 \\ 5/2 \\ 2 \end{array}\right]. \end{displaymath}


4.
Find the QR factorization of the matrix

\begin{displaymath}A=\left[\begin{array}{ccc}
1 & 4 & 5 \\
2 & 2 & 1 \\
2 & 5 & 1 \end{array}\right] \end{displaymath}


Answer:
Write ${\bf v}_1$, ${\bf v}_2$ and ${\bf v}_3$ for the columns of $A$. Put ${\bf b}_1={\bf v}_1$. Let

\begin{displaymath}{\bf b}_2={\bf v}_2-
\frac{{\bf v}_2\bullet {\bf b}_1}{{\bf ...
...ght]
=\left[\begin{array}{r} 2 \\ -2 \\ 1 \end{array}\right], \end{displaymath}

and

\begin{eqnarray*}{\bf b}_3
&=&{\bf v}_3
-\frac{{\bf v}_3\bullet{\bf b}_1}{{\b...
...right]=
\left[\begin{array}{r}2 \\ 1 \\ -2 \end{array}\right]. \end{eqnarray*}



Upon normalization, we get an ONB:

\begin{displaymath}{\bf u}_1=\frac{{\bf b}_1}{\Vert{\bf b}_1\Vert}
=\left[\begi...
...=\left[\begin{array}{r} 2/3 \\ 1/3 \\ -2/3 \end{array}\right]. \end{displaymath}

The $Q$ matrix in the $QR$-decomposition of $A$ is

\begin{displaymath}Q=\left[\begin{array}{rrr}
1/3 & 2/3 & 2/3 \\
2/3 &-2/3 & 1/3 \\
2/3 & 1/3 &-2/3 \end{array}\right] \end{displaymath}

and the $R$ matrix is

\begin{displaymath}R=Q^\top A=\left[\begin{array}{rrr}
1/3 & 2/3 & 2/3 \\
2/3...
...}
3 & 6 & 3 \\
0 & 3 & 3 \\
0 & 0 & 3 \end{array}\right]. \end{displaymath}




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C.K. Fong 2001-11-09