1. Help with Orthogonal Proof

Could anyone help me with the following proof?

Suppose $u,v\in V$. Prove that $=0$ if and only if $||u||\leq ||u+av||$ for all $a\in F$.

I already came up with a proof assuming that $=0$ and used the Pythagorean Theorem to prove that $||u||\leq ||u+av||$.
However, I don't know how to prove it the other way.

2. Originally Posted by zebra2147
Could anyone help me with the following proof?

Suppose $u,v\in V$. Prove that $=0$ if and only if $||u||\leq ||u+av||$ for all $a\in F$.

I already came up with a proof assuming that $=0$ and used the Pythagorean Theorem to prove that $||u||\leq ||u+av||$.
However, I don't know how to prove it the other way.
The proof will be slightly different depending on whether the field $F$ is the real or the complex numbers. Here's how to do the real case.

If $\|u\|\leqslant\|u+av\|$ then $\|u\|^2\leqslant\|u+av\|^2$. Write that as $\langle u,u\rangle \leqslant\langle u+av,u+av\rangle$, which simplifies to $a^2\|v\|^2 + 2a\langle u,v\rangle\geqslant0$. That is a quadratic in $a$ (with zero constant term), and if it is non-negative for all real $a$ then its discriminant cannot be positive. So $\langle u,v\rangle^2\leqslant0$, which obviously implies $\langle u,v\rangle=0$.

For the complex case, you have to feed a few complex conjugates into the calculation, but the method is essentially the same.

3. Is there anyway to show this without using the discriminate? My professor has made it pretty clear that we have not gone over discriminate so I'm probably not suppose to use that fact...

4. Originally Posted by zebra2147
Is there anyway to show this without using the discriminate? My professor has made it pretty clear that we have not gone over discriminate so I'm probably not suppose to use that fact...
You may be thinking of something else... the discriminant of the quadratic $ax^2+bx+c$ is simply $\sqrt{b^2-4ac}$, that's just college algebra.

If you don't want to do that though, I think this should work:

Begin as Opalg did.

$||u||\leq ||u+av||\Rightarrow ||u||^2\leq ||u+av||^2$
$\Rightarrow \leq $
$\Rightarrow \leq +2a+a^2$
$\Rightarrow a^2+2a\geq 0$
$\Rightarrow a\cdot (a+2)\geq 0$.

So we have a product of two terms must be nonnegative; it follows that each factor must have the same sign (both positive or both negative) (unless of course one or both is just 0). The expression

$a+2$

is simply a linear polynomial in $a$ (the inner product terms are really just constants). Because of what we said about whether this value is positive or negative, we get

$\mathrm{lim}_{a\rightarrow 0^+}(a+2)\geq 0$, and
$\mathrm{lim}_{a\rightarrow 0^-}(a+2)\leq 0$. (*)

But polynomials are continuous, so the limit from the left and right is the same:

$\mathrm{lim}_{a\rightarrow 0^+}(a+2)= img.top {vertical-align:15%;} $\mathrm{lim}_{a\rightarrow 0^-}(a+2)\geq 0$" alt="\mathrm{lim}_{a\rightarrow 0^+}(<v,v>a+2<u,v>)= $\mathrm{lim}_{a\rightarrow 0^-}(a+2)\geq 0$" />

But (*) tells us that this common quantity is both nonnegative and nonpositive; thus
$\mathrm{lim}_{a\rightarrow 0}(a+2)=0$.

So

$0=\mathrm{lim}_{a\rightarrow 0}(a+2)$
$=2\Rightarrow =0$.

Now, all this limit stuff may be too fancy; it's simply trying to say that a straight line satisfying the above must be 0 at the origin.

5. Another method, once you have got to the inequality $a^2\|v\|^2 + 2a\langle u,v\rangle\geqslant0$ (valid for all $a$), is to complete the square to get $\Bigl(a\|v\| + \frac{\langle u,v\rangle}{\|v\|}\Bigr)^2 - \frac{\langle u,v\rangle^2}{\|v\|^2}\geqslant0$.

Then if you put $a = -\langle u,v\rangle/\|v\|^2$, the inequality becomes $-\langle u,v\rangle^2/\|v\|^2\geqslant0$, which can only happen if $\langle u,v\rangle=0$.