Linear Algebra: proof 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. 
 A: 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.
A: 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.
A: 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.
A: 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:

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:

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.
