Proving $proj_{proj_{\vec u} \vec v} \vec v=proj_{\vec u} \vec v$

Question

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!

Answer 1

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

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

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}