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a)

$v_1$ is non-zero

Set $w_1$ = $v_1$

Define

$w_2$ = $v_2$ - ($\frac{w_1.v_2}{w_1.w_1}$)$w_1$

show that $w_1$ · $w_2$ = $0$

(b) Assume further that $w_2$ is non-zero.

Define

$w_3$ = $v_3$ - ($\frac{w_1.v_3}{w_1.w_1}$)$w_1$ - ($\frac{w_2.v_3}{w_2.w_2}$)$w_2$

show that $w_1$ · $w_3$ = $w_2$ · $w_3$ = $0$.


Have worked out that in a) $w_2$ = $v_2$-$v_2$ therefore =$0$ so $w_1$ · $w_2$ = $0$

Then for b) $w_3$ = $v_3$ - $v_3$ - $v_3$ therefore = -$v_3$

So how on earth do I then show that $w_1$ · $w_3$ = $w_2$ · $w_3$ = $0$.

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  • $\begingroup$ $w_3 \ne -v_3$ and it is to an extent irrelevant to the question. What is $w_1 \cdot w_3$? And what about $w_2 \cdot w_3$? $\endgroup$ – Anubis Black Jul 19 '17 at 14:49
  • $\begingroup$ $w_3$ must =$0$ and as they are parallel then their dot products must = $0$? Am I correct with part a? $\endgroup$ – Lyns Jul 19 '17 at 14:52
  • $\begingroup$ For part a) you must calculate $w_1 \cdot w_2 = w_1 \cdot v_2 - w_1 \cdot (...)w_1$. It's distributivity. $\endgroup$ – Anubis Black Jul 19 '17 at 14:57
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[Too long to be a comment. This is not intended for answering your question directly.]

You are doing some wrong algebra: the dot between two vectors means dot product.

The key word is "Gram-Schmidt process". Drawing a picture may be useful for seeing what is really going on.

enter image description here

enter image description here

You can figure out how the notations are corresponding to yours. In particularly, note that $$ proj_vw:=\frac{v\cdot w}{v\cdot v}v $$

The mentioned Wikipedia article above has an animation: enter image description here

See also these two YouTube videos:

Gram schmidt process in space

Gram schmidt process in plane

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Remark about your approach:

$w_2$ is not equal to $0$ in general, for example let $v_1=\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}$ and $v_2=\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$. Then $w_2=v_2 \neq 0$.

In general,$$\left( \frac{w_1.v_2}{w_1.w_1}\right)w_1 \neq v_2$$

Here is how I would approach part a,

\begin{align}w_1.w_2&=w_1.\left( v_2-\frac{w_1.v_2}{w_1.w_1}w_1\right) \\&=w_1.v_2-\frac{w_1.v_2}{w_1.w_1}(w_1.w_1)\\ &=w_1.v_2-w_1.v_2\\ &=0\end{align}

Hint: for part b, result from a is useful.

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  • $\begingroup$ \begin{align}w_1.w_3&=-\left( \frac{w_2.v_3}{w_2.w_2}\right)w_2.w_1 \\\end{align} \begin{align}w_2.w_3&=-\left( \frac{w_1.v_3}{w_1.w_1}\right)w_1.w_2 \\\end{align} $\endgroup$ – Lyns Jul 19 '17 at 15:51
  • $\begingroup$ recall that you know the value of $w_1.w_2$. $\endgroup$ – Siong Thye Goh Jul 19 '17 at 15:57
  • $\begingroup$ We have proven $w_1.w_2=0$ in part a, we can use the result directly in part b. $\endgroup$ – Siong Thye Goh Jul 19 '17 at 16:18
  • $\begingroup$ $0$ if I sub out $w_2$ one first one and $w_1$ on second. Then $w_1$.$w_2$=$( \frac{w_1.w_2.v_3^2}{w_3^2})$ which =$0$ $\endgroup$ – Lyns Jul 19 '17 at 16:22
  • $\begingroup$ Ok I see I am making a mountain out of it I didn't need to do that at all. Thank you for you patience. It's been a long day! $\endgroup$ – Lyns Jul 19 '17 at 16:42

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