Projection with an orthonormal basis I am studying for projection, but I have something not to understand.

Let $\{ q_1, q_2,\dots,q_m \}$ be an orthonormal basis for $\mathbb{C}^m$. Then, for an arbitrary vector $v\in \mathbb{C}^m$, $v$ can be decomposed into $$\sum_{i=1}^m (q_i^*v)q_i$$

I understood above part, but I didn't understand the next part :

$$v = \sum_{i=1}^{m}(q_i^*v)q_i = \sum_{i=1}^m (q_iq_i^*)v$$

How could change from $\displaystyle \sum_{i=1}^{m}(q_i^*v)q_i $ to  $\displaystyle \sum_{i=1}^{m}(q_iq_i^*)v$? 
Also, I couldn't understand what's the meaning of $q_iq_i^*$, cross product of an orthonormal basis. 
 A: Note that $q_i^* v$ is a scalar; it is the inner product of the vector $q_i$ and $v$.
We may interchange the order of a scalar $a$ and a vector $u$, i.e. $au = ua$. In this situation this gives
$$
   \sum_{i=1}^m (q_i^* v) q_i = \sum_{i=1}^m q_i (q_i^* v).
$$
By associativity of matrix multiplication, we then get
$$
 \sum_{i=1}^m q_i (q_i^* v) = \sum_{i=1}^m(q_i q_i^*)v.
$$
Note that the expression $q_i q_i^*$ is not an inner product, it is a $m \times m$ matrix.
What does this expression mean? If we take some orthonormal vectors $\{q_1, \ldots, q_k\}$, with $k < m$ (so not a basis of $\mathbb{C}^m$), then the matrix $\sum_{i=1}^k q_i q_i^*$ is the projection onto the space spanned by $\{q_1, \ldots, q_k\}$. It is illuminating to calculate this expression for $q_1 = (1\,0\, 0)^T$, $q_2=(0\, 1\, 0)^T$ in $\mathbb{C}^3$, the result is the matrix 
$$
 \sum_{i=1}^2 q_i q_i^* = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 &0 \end{pmatrix}.
$$ 
In the special case $k=m$, as in your question, the orthonormal basis spans the whole space $\mathbb{C}^m$ and thus $\sum_{i=1}^m q_i q_i^*$ is just the identity matrix.
A: It's not a direct formal manipulation, nevertheless both equations holds:
$$v=\sum_i(q_i^*v)q_i\\
v=\sum_i(q_iq_i^*)v$$
The second one is easy to check for $v=q_j$, and so by linearity, it holds for any vector $v$. 
