# length of orthogonal projection

I have a problem with the formula of orthonormal projection.

I know the orthogonal projection of a vector $v$ to one dimension subspace $U$ with basis $u$ is $$Pro^{v}_{U}=\dfrac{\langle v,u \rangle }{\langle u,u \rangle }u$$

but from the other hand the length of projection vector is inner product of $u$ and $v$, $\langle u,v \rangle .$

Now my problem is that if I want to find projection length from the first formula I have:

$$\langle Pro^{v}_{U},Pro^{v}_{U} \rangle =\langle \dfrac{\langle v,u \rangle }{\langle u,u \rangle }u,\dfrac{\langle v,u \rangle }{\langle u,u \rangle }u \rangle$$

which it is not equal to inner product of $u$ and $v$.

Where am I making mistake?

These 2 are not equal for example: $v=(2,3)$ , $u=(1,2)$

$$Pro^{v}_{U}=(\frac{8}{5},\frac{16}{5}),$$ $$\langle Pro^{v}_{U},Pro^{v}_{U} \rangle =\frac{320}{25},$$

but $\langle u,v \rangle =8.$

The lengh of projection vector is equal to $\langle u, v \rangle$ for $|u|=1$ and in this case
$$\mbox{Pro}^{v}_{U}= \langle v, u \rangle u$$
$$\langle v,u \rangle = \sqrt{\langle \mbox{Pro}^{v}_{U}, \mbox{Pro}^{v}_{U}\rangle} = \sqrt{\langle \langle v,u \rangle u,\langle v,u \rangle u \rangle}$$
• @shere where does the definition of lenght $<u,v>$ for inner product come from? there must be some sort of assumption or normalization – user Mar 15 '18 at 13:41