Orthogonal Projection in Arnoldi iteration. Let $H_n$ be n x n Hessenberg matrix, $K_n$ be nth order Krylov subspace, and $Q_n$ be the orthogonal matrix resulting from QR decomposition of $K_n$. Then, at the nth step of Arnoldi iteration, we get the following equation.
$$ H_n = Q_n^* AQ_n$$
I don't understand the following paragraph in the book, Numerical Linear Algebra, Trefethen.

"This matrix can be interpreted as the representation in the basis
  $\{q_1, ..., q_n\}$ of the orthogonal projection of A onto $K_n. $
  Consider the linear operator $K_n \rightarrow  K_n$ defined as follows: given $v \in K_n$, apply A to it, then orthogonally project $Av$ back into the space $K_n$."

I can't apprehend the two parts.
(1) In my thought, the orthogonal projection of A onto $K_n$ is $Q_nQ_n^*A$ because, in general, $Q_nQ_n^*$ is an orthogonal projection matrix wtih basis $\{q_1, ..., q_n\}$. But the above paragraph said that $Q_n^* AQ_n$ is the orthogonal projection of A onto $K_n$.
(2) The author said that $H_n$ is the linear operator such that for $v \in K_n$, apply A to it, then project $Av$ back into the space $K_n$. Howver, I think that given $v \in K_n$, first $v$ is applied to $Q_n$, followed by $A$. So, I think $AQ_nv$ is projected back into the space $K_n $. What's the wrong with my reasoning?
 A: It is true that $Q_nQ_n^*$ is the orthogonal projector into the span of $Q_n$, but this is in terms of the standard basis.
Perhaps the confusion in that the statement is about the projector in the basis of the $q_i$'s.
Suppose you know the representation of a vector $v$ in terms of the vectors in $Q_n$. That is, you know $c_i$ so that,
$$
v = \sum_{i=1}^N c_i q_i
$$
Then we can write $v = Q_nc$, which is the representation in the standard basis.
Now, as you noted, to project onto the span of $Q_n$ (which is the same as the span of $K_n$) we compute,
$$
Q_nQ_n^*Av 
= Q_nQ_n^*AQ_nc 
= Q_n(Q_n^*AQ_nc)
$$
The key is that the vector $Q_n^*AQ_nc$ gives the coefficients of $Q_nQ_n^*Av$ for a linear combination of the vectors in $Q_n$ (i.e. in the basis of the $q_i$'s.)
Thus, given the coefficients of some vector with respect to the basis of $q_i$'s (i.e. $c$ in this example), applying $Q_n^*AQ_n$ will give the coefficients of the projection onto the span of the $q_i$'s of that vector with respect to the basis of the $q_i$'s.
