# 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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