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

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

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You are on the right track. Try to prove the converse. –  anonymous Sep 4 '10 at 20:17
Your reasoning is perfectly fine as is stands. @Chandru1: Which converse? –  Rasmus Sep 4 '10 at 20:48
I do not think that AB is the standard way to denote the inner product of two vectors. For a more elegant proof, try to treat the vectors as they are, without decomposing the expression into coordinates. –  Tsuyoshi Ito Sep 4 '10 at 21:41
@Rasmus: I think @Chandru1 is referring to the other direction of the iff statement. But that is trivial: just go backwards in the original proof. No? –  Yaser Sulaiman Sep 4 '10 at 23:02

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...?

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