TITLE: Proving that $||A-B||=||A+B||\Leftrightarrow AB=0$
QUESTION [5 upvotes]: I have to prove that 
\begin{equation*}
||A-B||=||A+B||\Leftrightarrow AB=0
\end{equation*}
and I was wondering if this approach is correct, or if there's a better/more elegant way to prove this. 
Given n-dimensional vectors A and B, we can write $||A-B||=||A+B||$ as:  
\begin{equation*}
\sqrt{\sum\limits_{j=1}^{n}(a_j-b_j)^2}=\sqrt{\sum\limits_{j=1}^{n}(a_j+b_j)^2}.
\end{equation*}
Squaring both sides and expanding the binomials:
\begin{equation*}
\sum\limits_{j=1}^{n}a^2_j-2a_jb_j+b_j^2=\sum\limits_{j=1}^{n}a^2_j+2a_jb_j+b_j^2.
\end{equation*}
Simplifying:  
\begin{equation*}
-\sum\limits_{j=1}^{n}a_jb_j=\sum\limits_{j=1}^{n}a_jb_j,~\text{which holds true if and only if}~\sum\limits_{j=1}^{n}a_jb_j=0. 
\end{equation*}
Since $AB$ is equivalent to $\sum\limits_{j=1}^{n}a_jb_j$, then $||A-B||=||A+B||\Leftrightarrow AB=0$  
Thanks in advance.

REPLY [8 votes]: What you've done is correct, but I think it's better to work without coordinates; just with the definition of norm in terms of the dot product:
$$
\| A \| = +\sqrt{A\cdot A} \ .
$$
Then you may observe that, since $\|A \| \geq 0$,
$$
\|A+B\| =  \|A -B\| \  \Longleftrightarrow  \  \|A +B\|^2  =  \|A-B\|^2 .
$$
Now, for instance, compute the difference
\begin{align}
 \|A +B\|^2  -  \|A-B\|^2   &= (A+B)\cdot (A+B) - (A-B)\cdot(A-B)    \\
               &= A\cdot A + A\cdot B + B\cdot A + \cdots
\end{align}
EDIT. I forgot to point out an obvious geometric interpretation of this result: if you draw a parallelogram with sides $A$ and $B$, then $A+B$ and $A-B$ are the diagonals of the parallelogram, right? These diagonals are equal if and only if...?