# Show that $w_1$ · $w_3$ = $w_2$ · $w_3$ = $0$.

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 $w_1$ · $w_3$ = $w_2$ · $w_3$ = $0$.

• $w_3 \ne -v_3$ and it is to an extent irrelevant to the question. What is $w_1 \cdot w_3$? And what about $w_2 \cdot w_3$? – Anubis Black Jul 19 '17 at 14:49
• $w_3$ must =$0$ and as they are parallel then their dot products must = $0$? Am I correct with part a? – Lyns Jul 19 '17 at 14:52
• For part a) you must calculate $w_1 \cdot w_2 = w_1 \cdot v_2 - w_1 \cdot (...)w_1$. It's distributivity. – Anubis Black Jul 19 '17 at 14:57

[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.  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: Gram schmidt process in space

Gram schmidt process in plane

$w_2$ 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.

• \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} – Lyns Jul 19 '17 at 15:51
• recall that you know the value of $w_1.w_2$. – Siong Thye Goh Jul 19 '17 at 15:57
• We have proven $w_1.w_2=0$ in part a, we can use the result directly in part b. – Siong Thye Goh Jul 19 '17 at 16:18
• $0$ 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 =$0$ – Lyns Jul 19 '17 at 16:22
• 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! – Lyns Jul 19 '17 at 16:42