Linear Algebra: proof

Question

Show that $\alpha, \beta \in \mathbb R^n$ are orthogonal if, and only if, $\|\alpha-\beta\|=\|\alpha+\beta\|$

My first thought was to square each side as to get rid of the square root, and then its just a dot product but for some reason I am not reaching the desired result. Please help me.

Do you mean $n\times 1$ in exponent? Typeset as $n\times 1$ . (But I think the more usual notation for vectors is $\mathbb R^n$.) — Martin Sleziak, Apr 28 '13 at 15:23
Maybe this (and related question) might help you: From norm to scalar product. — Martin Sleziak, Apr 28 '13 at 15:25

score 4 · Answer 1 · answered Apr 28 '13 at 15:30

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Your first thought was correct. Square both side to get $\|\alpha-\beta\|=\|\alpha+\beta\|\iff \langle\alpha-\beta,\alpha-\beta\rangle=\langle\alpha+\beta,\alpha+\beta\rangle$.

Then use the bilinearity of the dot product to obtain $\langle\alpha-\beta,\alpha-\beta\rangle=\langle\alpha+\beta,\alpha+\beta\rangle\iff 4\langle\alpha,\beta\rangle=0.$

The last equation is equivalent to $\alpha,\beta$ being orthogonal.

answered Apr 28 '13 at 15:30

P..

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ok i see i should of kept going,i just was not sure if that was the right track. But since its an if and only if proof what is the other direction i need to go? – user74844 Apr 28 '13 at 15:40

score 2 · Answer 2 · answered Apr 28 '13 at 15:28

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You're on the right track, I'm not sure what the problem is!

Since both sides are positive, it is equivalent to the same equation squared. But then we can expand both squares to get $$ (x\pm y)\cdot(x\pm y) = x^2 +y^2 \pm 2 x\cdot y$$ Then take away the squared terms, and you're left with $4x\cdot y= 0$, your result.

answered Apr 28 '13 at 15:28

not all wrong

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ok that way makes sense, and once i did each dot product out i got the same result as you. But what is the other direction i need to go in order to finish the proof. – user74844 Apr 28 '13 at 15:46
The entire argument has $\iff$ symbols throughout it. – not all wrong Apr 28 '13 at 21:33

score 1 · Answer 3 · answered Apr 28 '13 at 15:30

1

Well, maybe not that neat, but still: $ \|\mathbf{x} - \mathbf{y}\| = \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$, $ \|\mathbf{x} +\mathbf{y}\| = \sqrt{\sum_{i=1}^n (x_i + y_i)^2}$, square them $\sum_{i=1}^n (x_i - y_i)^2=\sum_{i=1}^n (x_i + y_i)^2$ or $$\sum_{i=1}^n (x_i )^2+\sum_{i=1}^n (y_i)^2+2\sum_{i=1}^n x_i y_i=\sum_{i=1}^n (x_i )^2+\sum_{i=1}^n (y_i)^2-2\sum_{i=1}^n x_i y_i$$ or $$4\sum_{i=1}^n x_i y_i=4\mathbf{x}\cdot\mathbf{y}=0$$ which is exactly what you want.

answered Apr 28 '13 at 15:30

Caran-d'Ache

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I see the way you did it as well and it makes a lot of sense. But since its and if and only if proof, what is the other direction i need to go in order to finish the proof. – user74844 Apr 28 '13 at 15:41
@user74844 Well, you can reverse it. Say if $x,y\in \mathbb{R}^n$, $x$ and $y$ are orthigonal, then $\mathbf{x}\cdot\mathbf{y}=0$, so $4\mathbf{x}\cdot\mathbf{y}=0$. Just add and subtract $\mathbf{x}\cdot\mathbf{x}+\mathbf{y}\cdot\mathbf{y}$ and reorganize the terms to get what you want. – Caran-d'Ache Apr 28 '13 at 15:55
oh i see so fairly straight forward – user74844 Apr 28 '13 at 16:12

score 0 · Answer 4 · answered Aug 28 '13 at 12:16

We'll prove it for non-zero $\alpha$ and $\beta$ (the claim is straightforward otherwise).

We resolve $\beta$ into components $\beta^{\mathrm{para}}$ parallel to $\alpha$ and $\beta^{\mathrm{perp}}$ orthogonal to $\alpha$ (see the vector projection page on Wikipedia for details). This is depicted below:

$Resolving $\beta$ into $\beta^{\mathrm{para}}$ parallel to $\alpha$ and $\beta^{\mathrm{perp}}$ orthogonal to $\alpha$$

Then \begin{align*} ||\alpha+\beta|| &= ||\alpha+\beta^{\mathrm{para}}+\beta^{\mathrm{perp}}|| \\ &= \sqrt{||\alpha+\beta^{\mathrm{para}}||^2+||\beta^{\mathrm{perp}}||^2} \end{align*} by Pythagoras' Theorem, since $\alpha+\beta^{\mathrm{para}}$ and $\beta^{\mathrm{perp}}$ are orthogonal. Similarly \begin{align*} ||\alpha-\beta|| &= ||\alpha-\beta^{\mathrm{para}}-\beta^{\mathrm{perp}}|| \\ &= \sqrt{||\alpha-\beta^{\mathrm{para}}||^2+||\beta^{\mathrm{perp}}||^2}. \end{align*} This is depicted below:

$Identifying $\alpha+\beta$ and $\alpha-\beta$ in terms of $\alpha$, $\beta^{\mathrm{para}}$ and $\beta^{\mathrm{perp}}$$

Hence $||\alpha+\beta||=||\alpha-\beta||$ if and only if $$||\alpha+\beta^{\mathrm{para}}||=||\alpha-\beta^{\mathrm{para}}||$$ which happens if and only if $\beta^{\mathrm{para}}$ is the zero vector (since $\alpha$ is non-zero). This happens if and only if $\alpha$ and $\beta$ are orthogonal.

Linear Algebra: proof

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