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.

  • $\begingroup$ The LaTeX for scalar products is \langle and \rangle, as in $\langle v, w \rangle$. $\endgroup$ – Ryan Reich Apr 21 '14 at 4:56
  • $\begingroup$ @RyanReich I wont fix them but I will remember (the answer is a lot shorter than I thought), thanks. May I also ask how to write norms? (like: $||v||$) and underline things, or put them in bold (without \boldtext I think it is, 11 characters to make 1 bold is not worth it!) $\endgroup$ – Alec Teal Apr 21 '14 at 4:58
  • $\begingroup$ For norms: \lVert and \rVert, for bold: \textbf{} or \mathbf{} for text or for math. For underline: \underline{}, of course, though that one has some annoying problems. If you think the commands are too long, you can always make a shortcut: \newcommand*\norm[1]{\lvert#1\rvert}. $\endgroup$ – Ryan Reich Apr 21 '14 at 5:01
  • $\begingroup$ I fixed it for you. $\endgroup$ – Ryan Reich Apr 21 '14 at 5:08
  • $\begingroup$ @RyanReich thanks, you shouldn't have had to do that. $\endgroup$ – Alec Teal Apr 21 '14 at 12:43

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.

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  • $\begingroup$ That bit about "applying $<-,v_j>$ to (1), can you just do that? I'm guessing it's $<\text{LHS},v_j>=<\text{RHS},v_j>$ applied like that? $\endgroup$ – Alec Teal Apr 21 '14 at 4:55
  • $\begingroup$ @AlecTeal Yes exactly. Explicitly, $\langle \lambda_1 v_1+\dotsb+\lambda_n v_n,v_j\rangle=\langle \mathbf 0,v_j\rangle$ $\endgroup$ – Brian Fitzpatrick Apr 21 '14 at 7:21

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