I have been given by the question that the scalar product is positive
definite ($\ge 0$ always – and $\langle v,v\rangle >0 \text{ for } v\ne 0$) (is
there a LaTeX way for scalar products, that doesn't look
great). This is wrong, doesn't change anything though
If I use this I side-step something that throws me off, I want to do
the thing I'd happily avoid – because I'm avoiding it because I'm
uncertain (that's a weird sentence).
Anyway, assume $v_1,…,v_n$ are mutually perpendicular, non-zero
vectors (by this I mean $\langle v_i,v_j \rangle=0\iff i\ne j$). I
wish to prove they are linearly independent (it's surprising how far I
have I have gotten without using this theorem).
I went for contradiction, suppose they are linearly dependent, then
$\exists a_i \ne 0$ for $\sum a_iv_i=0$
Suppose without loss of generality (fancy way of saying "reorder to
make this true") $v_1\ne 0$, then:
$$v_1=\frac{-a_2}{a_1}v_2+…+\frac{-a_n}{a_1}v_n$$
Now consider $\langle v_1,v_2 \rangle$ (should I use $v_k$ with $k\ne
1$ rather than $v_2$? I'm proving by contradiction, any would work, I
chose 2 so I could use $…$s neatly.)
$$\langle v_1,v_2 \rangle
=\langle \frac{-a_2}{a_1}v_2+…+\frac{-a_n}{a_1}v_n,v_2 \rangle
=\frac{-a_2}{a_1}\langle v_2,v_2 \rangle+\cdots
+\frac{-a_n}{a_1}\langle v_n,v_2 \rangle$$
This is where I paused and chose to ask, if I use
positive-definite-ness(?) then:
$$\frac{-a_2}{a_1}\langle v_2,v_2 \rangle\le 0$$
(oh wait that's absolute rubbish, I was thinking "I cannot think of
any positive definite scalar products"!)
Back on track I want to say "by hypothesis $\langle v_k,v_2
\rangle=0$ for $k\ge 3$" then:
$$\langle v_1,v_2 \rangle=\frac{-a_2}{a_1}\langle v_2,v_2
\rangle+\cdots+\frac{-a_n}{a_1}\langle v_n,v_2 \rangle
=\frac{-a_2}{a_1}\langle v_2,v_2 \rangle$$
This is where I am stuck, I am not sure I can say "by hypothesis"
because I'm assuming (part of it) isn't true.
I'm also stuck because even if I use the positive-definite part to say
$\langle v_2,v_2 \rangle >0$ I don't know that $a_2\ne 0$. I'm not
very confident with proof by contradiction because I'm assuming the
thing I know about to be false.
I think that if things are lin. dependent (for non-zero vectors)
then there exists at least 2 coefficients that are non zero, one
to take the sum away from 0, and another dependent vector to bring it
back to zero. Geometrically I am not certain about this.
I'd love some help, I'm quite muddled.
Best Answer
Suppose $$ \lambda_1 v_1+\dotsb+\lambda_n v_n=\mathbf 0\tag{1} $$ Applying $\langle-,v_j\rangle$ to $(1)$ gives $$ \lambda_j\langle v_j,v_j\rangle=0\tag{2} $$ Moreover, $\langle v_j,v_j\rangle\neq0$ since $\langle v_k,v_l\rangle=0$ if and only if $k\neq l$. Thus $(2)$ implies $\lambda_j=0$. Hence $$ \lambda_1=\dotsb=\lambda_n=0 $$ so $v_1,\dotsc,v_n$ are linearly independent.