Proving $proj_{proj_{\vec u} \vec v} \vec v=proj_{\vec u} \vec v$ Can anyone show me how to prove: $proj_{proj_{\vec u} \vec v} \vec v=proj_{\vec u} \vec v$? I got confused trying to prove it (not geometrically)... 
Thanks in advance!
 A: Observe that $proj_{\vec u} \vec v$ is always of the form $c \vec u$ for some constant $c$. 
So it suffices to show that $proj_{\vec u} \vec v = proj_{\vec cu} \vec v$ for any nonzero constant $c$, which is pretty immediate from the definitions. (In the case where $c = 0$ your equation is not well-defined.)
A: Let $\langle \vec{a}|\vec{b} \rangle$ the inner product of two vectors $\vec{a}$ and $\vec{b}$. Perhaps this is not a notation familiar to you. But it is more organized when it is scalar which are large algebraic expressions by multiplying vectors. Recall that $\|a\|^2=\langle \vec{a}|\vec{a}\rangle$. Note that


*

*$\langle proj_{\vec u} (\vec v) | \vec v\rangle
= 
\left\langle 
\color{red}{\dfrac{\langle \vec u\big|\vec v\rangle }{\|\vec u\|^2}}\vec u \;\Bigg|\;\vec v \right\rangle
=
\color{red}{\dfrac{\langle \vec u\big|\vec v\rangle }{\|\vec u\|^2}}\left\langle \vec u\left|\right.\vec v\right\rangle
=
\color{blue}{\dfrac{\langle \vec u \left|\right. \vec v \rangle^2}{\|\vec u \|^2}}$

*$\|proj_{\vec u} (\vec v)\|^2
= 
\langle proj_{\vec u} (\vec v) | proj_{\vec u} (\vec v)\rangle
=
 \left\langle 
\color{red}{\dfrac{\langle \vec u\big|\vec v\rangle }{\|\vec u\|^2}}\vec u 
\;\Bigg|\; 
\color{red}{\dfrac{\langle \vec u\big|\vec v\rangle }{\|\vec u\|^2}}\vec u
\right\rangle
 \\ \hspace{7cm}
=\color{red}{\dfrac{\langle \vec u\big|\vec v\rangle }{\|\vec u\|^2}}\cdot \color{red}{\dfrac{\langle \vec u\big|\vec v\rangle }{\|\vec u\|^2}} \left\langle 
\vec u 
\left|\right. 
\vec u
\right\rangle 
=\color{blue}{\dfrac{\langle \vec u \left|\right. \vec v \rangle^2}{\|\vec u \|^2}}$
implies
\begin{align}
proj_{proj_{\vec u} \vec v} (\vec v)
=
&
\frac{\langle proj_{\vec u} (\vec v)\big| \vec v \rangle}{\|proj_{\vec u} (\vec v)\|^2} proj_{\vec u} (\vec v)
=
\frac{\color{blue}{\dfrac{\langle \vec u \left|\right. \vec v \rangle^2}{\|\vec u \|^2}}}
{\color{blue}{\dfrac{\langle \vec u \left|\right. \vec v \rangle^2}{\|\vec u \|^2}}}proj_{\vec u} (\vec v)
=proj_{\vec u} (\vec v)\\
\end{align}
