For nonzero vectors, how do you prove the following?
- $$\|u\|^2 = \|\text{projection of $u$ onto $v$}\|^2 + \|u - \text{projection of $u$ onto $v$}\|^2$$
I think what we need to do is split up the latter part into two, but can't we only do this if the vector "$u$" and the vector projection onto "$v$" are orthogonal? How do I go about proving this?
Hint: Think Pythagoras: the vectors $\mathrm{proj}_{\vec v}(\vec u)$ and $\vec u - \mathrm{proj}_{\vec v}(\vec u)$ are orthogonal, so the triangle whose vertices are $\vec 0$, $\vec u$ and $\mathrm{proj}_{\vec v}(\vec u)$ forms a right-angled triangle.
To verify that $\mathrm{proj}_{\vec v}(\vec u)$ and $\vec u - \mathrm{proj}_{\vec v}(\vec u)$ are orthogonal, note that $$\mathrm{proj}_{\vec v}(\vec u) = \frac{\vec u \cdot \vec v}{\vec v \cdot \vec v} \vec v$$ meaning that $$\mathrm{proj}_{\vec v}(\vec u) \cdot (\vec u - \mathrm{proj}_{\vec v}(\vec u)) = \left( \frac{\vec u \cdot \vec v}{\vec v \cdot \vec v} \vec v\ \cdot\ (\vec u - \frac{\vec u \cdot \vec v}{\vec v \cdot \vec v} \vec v)\right)\ \underbrace{=\ \cdots\ =}_{\text{you do this bit}}\ 0$$