I would like to prove the following equality between one projection in arbitrary basis $B=(e_{i})$ and the same projection in another basis $B'=(e'_{i})$ :
The projection projects a vector $\vec{w}$ onto $\vec{v}$ direction. $w$ and $v$ are expressed in $B$ basis whereas $w'$ and $v'$ are the same vectors but expressed in $B'$ basis.
Then, My goal is to get :
$$\text{proj}_{\vec{v}}(\vec{w}) = P\,\text{proj}_{\vec{v'}}(\vec{w'})\quad\quad(1)$$
with the matrix $P$ which is the transfer matrix between $B'$ and $B$ : so I have $\vec{v}=P\,\vec{v'}$, $\vec{v^{*}}=P^{-1}\,\vec{v'^{*}}$, $\vec{w}=P\,\vec{w'}$.
To start, I take the expression of $\text{proj}_{\vec{v}}(\vec{w})$ :
$$\text{proj}_{\vec{v}}(\vec{w})=vv^{*}w=P\,\vec{v'}\,P^{-1}\,\vec{v'^{*}}\, P\, \vec{w'}\quad\quad(2)$$
and from this, I would like to have :
$$\text{proj}_{\vec{v'}}(\vec{w'})=\vec{v'}\vec{v'^{*}}\vec{w'}$$
but I don't how to make disappear the matrices $P$ and $P^{-1}$ in equation $(2)$ since the product of matrices is not commutative.
If this was commutative, I could write from $(2)$ :
$$\text{proj}_{\vec{v}}(\vec{w})=vv^{*}w=P\,\vec{v'}\,P^{-1}\,P\,\vec{v'^{*}}\, \vec{w'}=P\,\vec{v'}\,\vec{v'^{*}}\, \vec{w'}=P\,\text{proj}_{\vec{v'}}(\vec{w'})\quad\quad(3)$$
Unfortunately, this swapping is not allowed.
How to circumvent this issue in order to get equation $(1)$ ?
EDIT 1:
Maybe I should write : $\vec{v^{*}}=\vec{v'^{*}}\,P^{-1}$ instead of $\vec{v^{*}}=P^{-1}\,\vec{v'^{*}}$, so it comes :
$$\text{proj}_{\vec{v}}(\vec{w})=vv^{*}w=P\,\vec{v'}\,\vec{v'^{*}}\,P^{-1}\,P\,\vec{w'}=P\,\vec{v'}\,\vec{v'^{*}}\,\vec{w'}=P\,\text{proj}_{\vec{v'}}(\vec{w'})\quad\quad(4)$$
By this way, I could find the simple expression between coordinates "$X$" of a vector expressed in "$B$" basis and its coordinates "$X'$" expressed in "$B'$" basis , like this :
$$X=PX'$$
Is equation(4) correct ?
EDIT 2: I have doubts about the following expression (see $(1))$ :
$$\bigg(\text{proj}_{\vec{v}}(\vec{w})\bigg)_{B} = P\,\bigg(\text{proj}_{\vec{v'}}(\vec{w'})\bigg)_{B'}$$
with $P=\text{Mat}_{BB'}$ the transfer matrix.
Why couldn't we remove the $P$ matrix and write simply :
$$\bigg(\text{proj}_{\vec{v}}(\vec{w})\bigg)_{B} = \bigg(\text{proj}_{\vec{v'}}(\vec{w'})\bigg)_{B'}\quad\quad(5)$$
It seems the projected vector into one basis $B$ is the same than projected vector but expressed into another basis $B'$.
Then, if I follow $(5)$, I get :
$$vv^{*}w=w^{i}\vec{e_i}=v'v'^{*}w'=w'^{j}\vec{e'_{j}}=w'^{j}P_{ij}\vec{e_i}$$
So one gets :
$w^{i}=w'^{j}P_{ij}$
This is the relation between contravariants components.
What expression is right :
$$\bigg(\text{proj}_{\vec{v}}(\vec{w})\bigg)_{B} = \bigg(\text{proj}_{\vec{v'}}(\vec{w'})\bigg)_{B'}\quad\quad(6)$$
OR
$$\bigg(\text{proj}_{\vec{v}}(\vec{w})\bigg)_{B} = P\, \bigg(\text{proj}_{\vec{v'}}(\vec{w'})\bigg)_{B'}\quad\quad(7)$$
???
EDIT 3:
Following the notations of @amd , my goal is not to write $$[\pi_v]_{\mathcal B}^{\mathcal B} = P [\pi_v]_{\mathcal B'}^{\mathcal B'}$$
but rather : $$[\pi_v w]_{\mathcal B} = P\,[\pi_v w]_{\mathcal B'}$$.
Indeed, we have $$<[v]_B,[w]_B> = <[v]_{B'},[w]_{B'}>$$ and $$[v]_{B}=P\,[v]_{B'}$$,
so finally we can write : $$[\pi_v w]_{\mathcal B}=P\,[\pi_v w]_{\mathcal B'}$$, is it right ?
With my notations, this would take the form :
$$\bigg(\text{proj}_{(\vec{v})_{B}}(\vec{w})_{B}\bigg)_{B} = P\,\bigg(\text{proj}_{(\vec{v})_{B'}}(\vec{w})_{B'})\bigg)_{B'}$$
wouldn't it ?
There’s not really anything tricky here. In particular, you don’t need to try so hard to eliminate the $P$’s in the resulting expression because they actually belong there. They’re essential to the inner product that’s inherent in the projection.
I’ll use somewhat different notation from that in the question because it’s important to distinguish the vectors themselves—elements of the inner product space $V$—from their coordinate tuples—elements of various copies of $\mathbb K^n$. The symbols $v$ and $w$ stand for vectors, i.e., $v,w\in V$. Their coordinate tuples relative to the ordered basis $\mathcal B$ are denoted $[v]_{\mathcal B}$ and $[w]_{\mathcal B}$, respectively. These are elements of $\mathbb K^n$. Similarly, if $L:V\to V$ is an automorphism of $V$, then its matrix relative to the “input” basis $\mathcal B'$ and “output” basis $\mathcal B$ is denoted by $[L]_{\mathcal B}^{\mathcal B'}$. This notation allows you to “cancel” a superscript against the subscript of the next term to its right in a multiplication. From this point of view, a change of basis doesn’t change the vector that the two coordinate tuples both represent, so a transfer matrix is a representation of the identity map $\operatorname{id}$.
The map $\pi_v:V\to V$ that orthogonally projects onto some fixed $v\in V$ is given by the formula $$\pi_v: w \mapsto {\langle w, v\rangle \over \langle v, v\rangle} v.\tag1$$ Here, $\langle\cdot,\cdot\rangle : V\times V\to\mathbb K$ stands for the inner product used to define orthogonality. Now, suppose that $\mathcal B$ is a basis in which $\langle w, v\rangle = [v]_{\mathcal B}^* [w]_{\mathcal B}$. Then, if $P=[\operatorname{id}]_{\mathcal B}^{\mathcal B'}$, we have $[v]_{\mathcal B}=P[v]_{\mathcal B'}$ and $[w]_{\mathcal B}=P[w]_{\mathcal B'}$, and so $$\langle w, v\rangle = \left(P[v]_{\mathcal B'}\right)^* \left(P[w]_{\mathcal B'}\right) = [v]_{\mathcal B'}^* P^*P [w]_{\mathcal B'}.\tag2$$ This is nothing more than an application of the change-of-basis formula for quadratic forms: if $Q$ is the matrix of a quadratic form relative to the basis $\mathcal B$ and $P = [\operatorname{id}]_{\mathcal B}^{\mathcal B'}$ the transfer matrix from another basis $\mathcal B'$, then the matrix of the quadratic form relative to $\mathcal B'$ is $P^*QP$. This is one of the ways in which you erred: $\langle w,v\rangle = [v]_{\mathcal B}^* [w]_{\mathcal B}$ for all $v$ and $w$ iff $\mathcal B$ is orthonormal. Otherwise, some positive-definite matrix other than the identity must come into play.
To reduce clutter, assume now that $\|v\|=1$. The derivation is similar, but messier, in the general case, for which there’ll be a $1/\langle v,v\rangle$ term floating around. I’m also assuming that $\mathcal B$ is orthonormal. For full generality, we’d have to start with $\langle w,v\rangle = [v]_{\mathcal B}^* Q [w]_{\mathcal B}$ for some positive-definite matrix $Q$. From (1) we have $$[\pi_v w]_{\mathcal B}^{\mathcal B} = [v]_{\mathcal B}^*[w]_{\mathcal B}[v]_{\mathcal B} = [v]_{\mathcal B} [v]_{\mathcal B}^*[w]_{\mathcal B},$$ so $[\pi_v]_{\mathcal B}^{\mathcal B} = [v]_{\mathcal B} [v]_{\mathcal B}^*$. Similarly, $$[\pi_v]_{\mathcal B'}^{\mathcal B'} = [v]_{\mathcal B'}^* P^*P [w]_{\mathcal B'} [v]_{\mathcal B'} = [v]_{\mathcal B'} [v]_{\mathcal B'}^* P^*P [w]_{\mathcal B'}$$ and $[\pi_v]_{\mathcal B'}^{\mathcal B'} = [v]_{\mathcal B'} [v]_{\mathcal B'}^* P^*P$. The two matrices are related by the usual change-of-basis formula: $$[\pi_v]_{\mathcal B}^{\mathcal B} = [\operatorname{id}]_{\mathcal B}^{\mathcal B'}[\pi_v]_{\mathcal B'}^{\mathcal B'} [\operatorname{id}]_{\mathcal B'}^{\mathcal B} = P[\pi_v]_{\mathcal B'}^{\mathcal B'} P^{-1}.$$ On the other hand, $$[\pi_v]_{\mathcal B}^{\mathcal B} = [v]_{\mathcal B} [v]_{\mathcal B}^* = \left(P [v]_{\mathcal B'}\right) \left(P [v]_{\mathcal B'}\right)^* = P [v]_{\mathcal B'} [v]_{\mathcal B'}^* P^* = P [\pi_v]_{\mathcal B'}^{\mathcal B'} P^{-1}.$$ If $\mathcal B'$ is also orthonormal, then $P^*P = I$ and this term will vanish, as you were trying to make it do, but you still won’t have $[\pi_v]_{\mathcal B}^{\mathcal B} = P [\pi_v]_{\mathcal B'}^{\mathcal B'}$ because of what’s effectively a type mismatch: the left-hand side expects its inputs expressed relative to $\mathcal B$, but the right-hand side expects inputs relative to $\mathcal B'$.