Proving that $||A-B||=||A+B||\Leftrightarrow AB=0$

Question

I have to prove that $||A-B||=||A+B||\Leftrightarrow AB=0$ 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:

$\sqrt{\sum\limits_{j=1}^{n}(a_j-b_j)^2}=\sqrt{\sum\limits_{j=1}^{n}(a_j+b_j)^2}$

Squaring both sides and expanding the binomials:

$\sum\limits_{j=1}^{n}a^2_j-2a_jb_j+b_j^2=\sum\limits_{j=1}^{n}a^2_j+2a_jb_j+b_j^2$

Simplifying:

$-\sum\limits_{j=1}^{n}a_jb_j=\sum\limits_{j=1}^{n}a_jb_j$, which holds true if and only if $\sum\limits_{j=1}^{n}a_jb_j=0$

Since $AB$ is equivalent to $\sum\limits_{j=1}^{n}a_jb_j$, then $||A-B||=||A+B||\Leftrightarrow AB=0$

Thanks in advance.

Answer 1

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...?