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# The inner product of two collinear vectors is the product of their norms

Theorem : Iff two vectors $u,v$ are collinear of the same direction ($u = \lambda v$, $\lambda > 0$), their inner product will be $\langle u, v \rangle = \| u \| \| v \|$

Proof : If $u = \lambda v$, then

\begin{eqnarray} \langle u, v \rangle - \| u \| \| v \| &=& \langle \lambda v, v \rangle - \sqrt{\langle u , u \rangle \langle v , v \rangle}\\ &=& \lambda \langle v, v \rangle - \sqrt{\langle \lambda v , \lambda v \rangle \langle v , v \rangle}\\ &=& |\lambda| \| v \|^2 - |\lambda| \sqrt{\langle v , v \rangle \langle v , v \rangle} \\ &=& |\lambda| \| v \|^2 - |\lambda| \| v \|^2 \\ &=& 0 \end{eqnarray}

Conversely, if $\langle u, v \rangle = \| u \| \| v \|$, consider a Gram-Schmidt orthonormal basis constructed from $u$, in which case $v = \lambda u + v^i e_i$. Then we get

\begin{eqnarray} \langle u, v \rangle - \| u \| \| v \| &=& \langle u, \lambda u + v^i e_i \rangle - \| u \| \| \lambda u + v^i e_i \|\\ &=& \lambda \langle u, u \rangle - \| u \| \| \lambda u + v^i e_i \|\\ &=& \| u \| (\lambda \| u \| - \| \lambda u + v^i e_i \|)\\ &=& \| u \| (\lambda \| u \| - \sqrt{ \| \lambda u\|^2 + \| v^i e_i \|^2 }) \end{eqnarray}

For any $v \neq 0$, the square root is strictly larger than $\lambda \| u \|$, and therefore not equal to zero. Therefore, $v = 0$ and

\begin{eqnarray} \| u \| (\lambda \| u \| - \sqrt{ \| \lambda u\|^2}) &=& \| u \| (\lambda \| u \| - \| \lambda u\|)\\ &=& \| u \| (\lambda \| u \| - |\lambda| \| u\|)\\ &=& \| u \|^2 (\lambda - |\lambda|) \end{eqnarray}

This will only be zero if $\lambda$ is positive.