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

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

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

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

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

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

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

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

4.

Find the QR factorization of the matrix

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

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

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

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:

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

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

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

and the $R$ matrix is

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