From this link : [http://math.ucr.edu/home/baez/lie/node12.html][1], it is said that, from starting a quantum state, $$\vec{v}=a|\text{up} \rangle+b|\text{down} \rangle,$$ we can define the matrix projection by doing:
$$\text{Matrix}_\text{projection}=vv^{*} =\left[ \begin{array}{cc}a \\ b \end{array} \right] \left[ \begin{array}{cc}a^* & b^* \end{array} \right] = \left[ \begin{array}{cc} a a^* & a b^* \\ a^* b & b b^* \end{array} \right] \, .$$
From now, I will write " $v$ " instead of " $\vec{v}$ " for abreviation of vectors (except for basis vectors $\vec{v_i}$, see at the end).
The matrix above looks like a tensorial product between vector $v$ and its dual $v^{*}$.
Then, one introduces into this link the projection of an arbitrary vector $w$ along vector $v$ like this : $$\text{proj}_v(w) = \langle v, w \rangle v$$ I don't understand the following equation: $$v (v^* w) = (v v^*)w \, $$
Indeed, I am stuck on the transition
$$v(v^* w) = vv^* w$$
because "$vv^* w=\text{Matrix}_\text{projection}\,w$" does not make appear the vectors basis $\vec{v_i}$ unlike the term $$v(v^{*}w)=\sum_{i}(v_{i}^{*}w) \vec{v_{i}}$$
$vv^{*} w$ represents the action of projection matrix on $w$ vector but doesn't specify the basis vector $\vec{v_{i}}$.
I would like to get clarifications about this issue (maybe notations I use or used are bad since I prefer considering coordinates into basis vectors).
Elements of answer :
I try to give some elements for my original issue.
1)For the first expression : If I take the expression $(\vec{v}.\vec{v}^{*})$, I can assimilate it to a (1,1) tensor, so like a matrix equal to in my case :
$$\begin{pmatrix} v^{1}_{1} & v^{1}_{2}\\ v^{2}_{1} & v^{2}_{2}\\ \end{pmatrix}$$
So, I can write the $i-th$ component, I get :
$$\bigg[(\vec{v}\vec{v^*}) \vec{w}\bigg]_{i}=\sum_{j} v^{i}_{j} w^{j}$$
2) For the second expression : $\vec{v}(\vec{v^*} \vec{w})$, one can have with :
$$\vec{v}=\sum_{i}y^{i}\vec{e_{i}}$$
So by taking the $i-th$ component of this expression (i.e relatively to $\vec{e_{i}}$ basis vector) :
$$\bigg[\vec{v}(\vec{v^*} \vec{w})\bigg]_{i}=\sum_{j}x_{j}y^{i}\,w^{j}=\sum_{j}v^{i}_{j}\,w^{j}$$
with $x_{j}y^{i} = v^{i}_{j}$
3) Conclusion : Finally, I find the same expressions for the $i-th$ component of output vector (with first and second expression, i.e $$(\vec{v}\vec{v^*}) \vec{w}=(\vec{v^{*}}\vec{w}) \vec{v}$$
But my confusion comes from the case 1), especially the equality :
$$(\vec{v}\vec{v^*}) \vec{w}=\sum_{i} \sum_{j} v^{i}_{j} \sum_{k} w^{k}\vec{e_{k}}$$
I should rather introduce the $\vec{e_{i}}\otimes\vec{e^{j}}$ :
$$(\vec{v}\vec{v^*}) \vec{w}=\sum_{i} \sum_{j} v^{i}_{j} \vec{e_{i}}\otimes\vec{e^{j}} \sum_{k} w^{k}\vec{e_{k}}$$
but this introduction of term $\vec{e_{i}}\otimes\vec{e^{j}}$ doesn't seem to be useful, this even doesn't change nothing in my demo.
From the begining, I tried to have a matricial product point of view for both expression, so I would like to know if I can extend the product $(\vec{v}\vec{v^*})$ as a tensorial product of vector $\vec{v}$ and its dual such as I can manipulate it like a classical matrix operating on vector $\vec{w}$ ?
Anyone could confirm me this implicit extension in expression ($\vec{v}\vec{v^{*}})$ since this product doesn't look like naturally to a matrix ?
