Issue with canonical proof of Schwartz inequality So in almost all recounts of the proof, we just "define" $$\left|Z\right\rangle = \left|V\right\rangle - 
\frac{\left\langle \ W| \ V\right\rangle }{\mid W\mid^2}\left|W\right\rangle $$
to start the proof, however I am not sure what the motivation for this form is.
Here, it is motivated by looking at the projection/rejection of one state onto the other, but I do not understand why the rejection of $∣a⟩$ from $∣b⟩$ is defined as $∣b–αa⟩$?
So my questions are:
How does one motivate the form for $|z\rangle$?
why is the rejection of $∣a⟩$ from $∣b⟩$ defined as $∣b–αa⟩$?
 A: The basic idea is that a vector lets you split the space into a direct sum of a one-dimensional subspace it spans and an orthogonal complement. The decomposition you ask about is exactly this decomposition. Let $|v\rangle$ and $|w\rangle$ be two vectors in some inner product space ${\cal H}$ (it could be a Hilbert space or not, it doesn't matter for the following discussion). If you just focus on $|w\rangle$ it will span a one-dimensional subspace, namely:
$$W\equiv \{\alpha |w\rangle \in{\cal H} : \alpha\in \mathbb{C}\}.$$
Now, ${\cal H}$ may then be decomposed in a direct sum ${\cal H}=W\oplus W^\perp$ where $W^\perp$ is the orthogonal complement, defined to be $$W^\perp \equiv \{|u\rangle \in{\cal H} : \langle w|u\rangle=0\}.$$
We can show this explicitly and in the process the definition you ask about will show up naturally. Indeed, showing that ${\cal H}=W\oplus W^\perp$ means finding unique $\alpha |w\rangle\in W$ and $|u\rangle\in W^\perp$ such that $$|v\rangle=\alpha |w\rangle+|u\rangle.$$
We may immediately determine those two elements by taking the inner product with $|w\rangle$. Indeed, in doing so we first determine $\alpha$ and hence the component along $W$:$$\langle w|v\rangle=\alpha \langle w|w\rangle+\langle w|u\rangle=\alpha\langle w|w\rangle\Longrightarrow \alpha = \dfrac{\langle w|v\rangle}{\langle w|w\rangle}.$$
Then since we want $|v\rangle=\alpha|w\rangle+|u\rangle$, where $\alpha$ has already been determined, subtracting $\alpha|w\rangle$ we have just found the $W^\perp$ component: $$|u\rangle=|v\rangle-\dfrac{\langle w|v\rangle}{\langle w|w\rangle}|w\rangle.$$
Therein appears the definition you seek to motivate. It is the part of $|v\rangle$ living in the orthogonal complement of $|w\rangle$.
A: Let a unit vector $\;\mathbf{n}=(\rm n_1,n_2,n_3)\,, \Vert\mathbf{n}\Vert=1$. Any vector $\;\mathbf{r}\;$ could be decomposed in two components with respect to $\;\mathbf{n}\;$, see Figure-01 in the bottom
\begin{equation}
   \mathbf{r}\boldsymbol{=}\mathbf{r}_\|\boldsymbol{+}\mathbf{r}_\bot
   \tag{01}\label{01}
\end{equation}
one parallel and the other normal to axis $\mathbf{n}$ respectively
\begin{align}
   \mathbf{r}_\| &\boldsymbol{=}\left(\mathbf{n}\boldsymbol{\cdot}\mathbf{r}\right)\mathbf{n}
   \tag{02.1}\label{02.1}\\
   \mathbf{r}_\bot &\boldsymbol{=}\left(\mathbf{n}\boldsymbol{\times}\mathbf{r}\right)\boldsymbol{\times}\mathbf{n}= \mathbf{r}\boldsymbol{-}(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}
   \tag{02.2}\label{02.2}
\end{align}
that is
\begin{equation}   \mathbf{r}\boldsymbol{=}\left(\mathbf{n}\boldsymbol{\cdot}\mathbf{r}\right)\mathbf{n}\boldsymbol{+}\left(\mathbf{n}\boldsymbol{\times}\mathbf{r}\right)\boldsymbol{\times} \mathbf{n}
   \tag{03}\label{03}
\end{equation}
If we want to decompose a vector $\;\mathbf{r}\;$ with respect to a vector $\;\mathbf{w}\;$ not necessarily of unit norm, then we produce the unit vector
\begin{equation}
\mathbf{n}\boldsymbol{=}\dfrac{\mathbf{w}}{\Vert\mathbf{w}\Vert}\boldsymbol{=}\dfrac{\mathbf{w}}{\mathrm w}
   \tag{04}\label{04}
\end{equation}
and the expressions \eqref{02.1},\eqref{02.2} take the form
\begin{align}
   \mathbf{r}_\| &\boldsymbol{=}\left(\mathbf{n}\boldsymbol{\cdot}\mathbf{r}\right)\mathbf{n}\boldsymbol{=}\dfrac{\left(\mathbf{w}\boldsymbol{\cdot}\mathbf{r}\right)}{\Vert\mathbf{w}\Vert^2}\mathbf{w}
   \tag{05.1}\label{05.1}\\
   \mathbf{r}_\bot &\boldsymbol{=}\left(\mathbf{n}\boldsymbol{\times}\mathbf{r}\right)\boldsymbol{\times}\mathbf{n}\boldsymbol{=}\mathbf{r}\boldsymbol{-}(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}\boldsymbol{=}\mathbf{r}\boldsymbol{-}\dfrac{\left(\mathbf{w}\boldsymbol{\cdot}\mathbf{r}\right)}{\Vert\mathbf{w}\Vert^2}\mathbf{w}
   \tag{05.2}\label{05.2}
\end{align}


