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So for real-valued vectors the statement

$$ \lVert u+v\rVert = \lVert u-v\rVert $$

holds if and only if $$ \langle u,v\rangle =0 $$

I want to state this now instead for complex-valued vectors

$$\lVert u+v\rVert =\lVert u-v\rVert \iff \langle u,v\rangle=-\langle u,v\rangle $$

Im not seeing the connection, but is the right hand side equal to the conjugate of $\langle u,v\rangle$ or?

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    by real valued/complex valued vectors you mean vectors in $\mathbb{C}^n$? – 2017-02-10
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    Not sure, they only say that I should re-state the statement when the inner product is complex.. – 2017-02-10
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    if you are working on an inner product space and with a norm induced from them, you can do $\|u\pm v\|=\langle u\pm v,u\pm v\rangle=\langle u,u\rangle + \langle u,\pm v\rangle + \langle \pm v,u \rangle + \langle \pm v,\pm v\rangle$, now you can use conjugate symmtry,linearity etc – 2017-02-10
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    And they actually didnt write real valued vectors, just that for a real inner product we have that : $$ ||u+v|| = ||u-v|| \iff =0$$ re-state it for a complex inner product.. – 2017-02-10
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    What I deduce from $$||u+v|| = ||u-v||$$ was that $$=-$$ But this only holds if =0 for a real inner product. But for the complex inner product im not so sure if there is something else that can be done with that line? – 2017-02-10
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    You got your condition wrong; it should be $\langle u,v \rangle = -\langle v,u\rangle$ – 2017-02-10
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    But isnt $$||u+v|| = \sqrt{ ||u||^{2} + ||v||^{2} + 2}$$ and $$ ||u-v|| = \sqrt{ ||u||^{2} + ||v||^{2} - 2} $$? – 2017-02-10
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    Not really, because $\langle u,v\rangle =\overline{\langle u,v\rangle}$. You need conjugation, so what you wrote there is only true if $\langle u,v\rangle$ is real – 2017-02-10
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    Yeah I meant to write down my condition for the real inner product ... And in some way make use of it for complex inner product statement.. – 2017-02-10

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Use the polarization identities (since you're in a real vector space):

$$\langle u,v\rangle = {1\over 4}(\lVert u+v\rVert^2-\lVert u-v\rVert^2).$$

For complex:

$$\langle u,v\rangle = {1\over 4}(\lVert u+v\rVert^2-\lVert u-v\rVert^2+i\lVert u+iv\rVert^2 -i\lVert u-iv\rVert^2).$$

If all they want is a restatement then just write it out:

$$\begin{cases}\lVert u+v\rVert^2=\langle u+v,u+v\rangle=\lVert u\rVert^2+\lVert v\rVert^2+\langle u,v\rangle +\langle v,u\rangle \\ \lVert u-v\rVert^2=\langle u-v,u-v\rangle=\lVert u\rVert^2+\lVert v\rVert^2-\langle u,v\rangle -\langle v,u\rangle\end{cases}$$

Then equality means

$$\langle u,v\rangle +\overline{\langle u,v\rangle} = -(\langle u,v\rangle+\overline{\langle u,v\rangle})$$

So $\text{Re}(\langle u,v\rangle)=0$ is the condition.

You can even verify this in the basic case of $V=\Bbb C$ where the inner product is given as $\langle u,v\rangle= u\overline{v}$. Let $u=1, v=i$ then $\langle u,v\rangle =(1)(-i)$ and clearly $\lVert u+v\rVert^2 = \lVert u-v\rVert^2=2$.

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    I just want to re-state the first statement for complex valued vectors...was my idea wrong or? – 2017-02-10
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    @fejz1234 nearly, check my edit for the full formula. – 2017-02-10
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    We haven't had the polarization identity in this course,not sure about this...is there no other general way to solving it? – 2017-02-10
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    where did $x,y$ pop up from? – 2017-02-10
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    @user160738 haha, oops, good call, I'll fix that. X_X – 2017-02-10
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    @fejz1234 remember not all results are one-step, you can go about establishing the polarization identity as a lemma, it's easy from the properties of the inner product. – 2017-02-10
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    @fejz1234 well if that's all you're looking for, it's simple. Check the edit. – 2017-02-10
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    Oh okay buy why are we squaring the norm? – 2017-02-10
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    @fejz1234 because weird square roots only make things harder. Since $\lVert x\rVert\ge 0$ you know $\lVert x\rVert =\lVert y\rVert \iff \lVert x\rVert^2=\lVert y\rVert^2$ and the latter more naturally shows up in inner product relations to norms. If you wish to make things harder on yourself, you may also put square roots everywhere, but I do not recommend it. – 2017-02-10
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    Okay sounds reasonable...silly question but how did you deduce the condition that Re() = 0.. im not used to complex inner product.. – 2017-02-10
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    @fejz1234 This is just a fact about complex numbers, $z+\overline{z} = 2 \text{Re}(z)$. But $2\text{Re}(z) = 0\iff \text{Re}(z) = 0$, so I thought the latter was a nicer conclusion to state. – 2017-02-10
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    So for that equality to hold the Re(z) has to be 0, but nothing is said about the Im(z) or? – 2017-02-10
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    @fejz1234 yes, there is nothing else in the statement, it can be anything. – 2017-02-10
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    But how can the equality hold if you have that Im(z) = -Im(z) ? – 2017-02-10
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    @fejz1234 $\text{Im}(z)$ has nothing to do with this. $\text{Im}(z)$ and $\text{Re}(z)$ do not interact, they can vary independently of one another. Look over the problem again, nowhere does $\text{Im}(z)$ make an appearance, in particular nowhere do I come up with $\text{Im}(z) = -\text{Im}(z)$. – 2017-02-10
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    So the equality can simplify to $$2 = -2$$ (since the conjugate of inner product is equal to the inner product? ) But this is basically the same conclusion as when the inner product is real too right? – 2017-02-10
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    @fejz1234 no, not at all look at my example $u=1, v=i$ then $\langle u, v\rangle = -i$ and clearly $-2i\ne 2i$. Remember $\langle u,v\rangle\ne \langle v,u\rangle$ all the time in the complex case (which is the one you care about). In the real case $\langle u,v\rangle \in\Bbb R$ so the condition $2\text{Re}(\langle u,v\rangle)=0\iff \langle u,v\rangle=0$, but in the complex case these are no longer equivalent. – 2017-02-10
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    Does =-i , because you conjugate the second argument or? – 2017-02-10
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    @fejz1234 yes, that's what is written. – 2017-02-10
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    Okay so I think im getting a better grasp of this.. But am I right when I say that the equality you wrote down can be simplified to $$2=-2$$? Since the complex inner product of 2 vectors is equal to the conjugate of the inner product of the same 2 vectors? – 2017-02-10
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    @fejz1234 No, it becomes $2\langle u,v\rangle = -2\overline{\langle u, v\rangle}$ which is different, since your condition would imply $\langle u,v\rangle=0$ which is not necessarily true, only the real part needs to be $0$. You could also write this as $\langle u,v\rangle = -\langle v,u\rangle$. – 2017-02-10
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    But why does the right hand side get conjugated? We had the same thing in both sides, just with different signs? im confused.. – 2017-02-10
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    @fejz1234 that's one of the properties of a complex inner-product. The only reason it **looks** like $\langle u,v\rangle =\langle v,u\rangle$ in the real case is because real numbers are their own complex conjugates. – 2017-02-10
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    Okay ...but hasnt our vectors u and v also an imaginary part? u=a+bi,v=c+di Why do we have no conditions on the imaginary part of the vectors, but only on the real? – 2017-02-10
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    @fejz1234 this has nothing to do with the vectors themselves, this is about their inner product. Remember $\langle \cdot,\cdot\rangle$ is a map which takes two vectors and spits out a complex number. It is that number we are talking about. And my counter question to you is why **should** we have any conditions on the imaginary part? The math shows us there is **only** a condition on the real part, so that's the way it is. Remember, just because something has multiple parts, doesn't mean all of them come into play all the time. In this case, only the real part is relevant, and that's fine. – 2017-02-10
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    Yea true, but the complex number that it spits out..isnt that on the form z=a+bi ? What made me confused was that even if a= 0 maybe the imaginary parts of both sides of the equality doesnt match.. but maybe im over-complicating this exercise hehe – 2017-02-10
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    @fejz1234 indeed you are, the imaginary parts might not match, but **it doesn't matter**, that's the beauty, we know that the two norms are equal iff the real parts are zero, that is **all that matters**, that's what we proved. – 2017-02-10
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    Okay sounds reasonable thanks for your time hehe ;) – 2017-02-10