Orthogonal Projection in Arnoldi iteration.

by alryosha   Last Updated October 10, 2019 03:20 AM - source

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.

"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?



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