Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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*}


\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.

share|cite|improve this question
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
up vote 8 down vote accepted

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...?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.