$1$: $($Section $3\mathrm{.}4$: Modified Gram$-$Schmidt; Exercise $3\mathrm{.}4\mathrm{.}26)$ Consider the

following three linearly independent vectors for some $>0$ :

v$1=$

$$

$$

$$

$$

$1$

$0$

$0$

$$

$$

$$

$,$ v$2=$

$$

$$

$$

$$

$1$

$0$

$0$

$$

$$

$$

$,$ v$3=$

$$

$$

$$

$$

$1$

$0$

$0$

$$

$$

$$

$$

As gets smaller, these vectors start to approach being linearly dependent. Let

us examine what happens when $>0$ is very small, so that $2<$ u$,$ where u unit

round$-$off on our computer. In floating point arithmetic, our computer would see

$1+$ $2=1.$

$($a$)$ Use the classical Gram$-$Schmidt algorithm to orthogonalize $\{$v$1,$ v$2,$ v$3\}$ to

produce $\{$q$1,$ q$2,$ q$3\},$ assuming that anytime $2$ is encountered it becomes zero, as

a model of rounding error. Calculate $($q$1,$ q$2)2$ and confirm it is $$$/2,$ which is

nearly zero for small $.$ But then calculate $($q$2,$ q$3)2$ and confirm it is $1/2,$ which is

far from zero.

$($b$)$ Use the modified Gram$-$Schmidt algorithm to orthogonalize $\{$v$1,$ v$2,$ v$3\}$ to

produce $\{$q$1,$ q$2,$ q$3\},$ assuming again that $2$ evaluates to zero. Calculation of q$2$

and hence $($q$1,$ q$2)2$ will be unchanged, but q$3$ is modified: Confirm that $($q$2,$ q$3)2$

now evaluates to zero. Confirm also that $($q$1,$ q$3)2$ evaluates to $$$/6,$ also nearly

zero.

$($Hint: Just compute three steps in both cases; calculation will simplify due to

$1+$ $2=1.)$

$1$