find a scalar $\alpha$ that satisfy the following

Question

when $v\neq 0$, find a scalar $\alpha$ such that $z:=u-\alpha v$ satisfies $\left \langle z,v \right \rangle = 0$

Is there some trick to this? I tried solving this explicitly and I just ended up with the following equation and got stuck.

$\left \langle u,v \right \rangle = \alpha\left \langle v,v \right \rangle$

Any help or insight is deeply appreciated

Answer 1

If $\langle u , v \rangle = \alpha \langle v, v \rangle$, then…

\begin{equation*}\boxed{ \alpha = \dfrac{\langle u , v \rangle}{\langle v , v \rangle} \left(=\dfrac{\langle u, v \rangle}{\|v\|^2}\right) }\end{equation*}

If you find that surprising, take a moment to think back to the "vectors" in geometry/physics, and finding the component of one vector along the direction of another.

Answer 2

So indeed we have $$\langle u,v \rangle = \alpha \langle v,v \rangle$$

All three of this numbers are scalars, and $v \neq 0$ implies $\langle v,v \rangle > 0$ as well, by positive-definitess of the scalar product.

So we can just define $$\alpha = \frac{\langle u,v \rangle}{\langle v,v \rangle}$$

and we are done:

$$\langle z,v \rangle = \langle u - \alpha v,v \rangle = \langle u,v \rangle - \alpha \langle v,v \rangle = \langle u,v \rangle - \langle u,v \rangle = 0$$

as required.