Let
- $n\in\mathbb N$
- $A\in\mathbb C^{n\times n}$ be invertible
- $b\in\mathbb C^n$
- $x_0\in\mathbb C^n$ and $$r_0:=b-Ax_0$$
Moreover, for all $i\in\mathbb N$, let $$\mathcal K_i:=\operatorname{span}\left\{A^0r_0,\ldots,A^{i-1}r_0\right\}\;,$$ $\left\{q_1,\ldots,q_i\right\}$ be an orthonormal basis of $\mathcal K_i$ and $Q_i$ be the matrix whose $k$th column is $q_k$. Now, again for all $i\in\mathbb N$, let $$h_{ki}:=\langle Aq_i,q_k\rangle\;\;\;\text{for }k\in\left\{1,\ldots,i\right\}\;,$$ $h_{(i+1)i}:=\left\|\tilde q_{i+1}\right\|$ with $$\tilde q_{i+1}:=Aq_i-\sum_{k=1}^ih_{ki}$$ and $H_i$ denote the element of $\mathbb C^{(i+1)\times i}$ whose $k$th column is $$\left(\begin{matrix}h_{1k}\\\vdots\\h_{(k+1)k}\\0\\\vdots\\0\end{matrix}\right)\;.\tag1$$ Furthermore, let $$H_i=U_iR_i\;,\tag2$$ with $U_i\in\mathbb C^{(i+1)\times i}$ being unitary and $R_i\in\mathbb C^{i\times i}$ being upper triangular, be the QR-decomposition of $H_i$.
Let $i\in\mathbb N$. It's easy to see that $$y_i=\underset{y\in\mathcal K_i}{\operatorname{arg min}}\left\|Ay-r_0\right\|\Leftrightarrow y_i=Q_i\lambda_i\text{ with }\lambda_i=\underset{\lambda\in\mathbb C^i}{\operatorname{arg min}\left\|r_0-AQ_i\lambda\right\|}\tag3\;.$$ How can we show that if $\lambda_i\in\mathbb C^i$ is a solution of $$R_i\lambda_i=\left\|r_0\right\|U_i^\ast e_1=:z_i\tag4\;,$$ with $e_1=(1,0,\ldots,0)^T\in\mathbb C^{i+1}$, $\lambda_i$ is a solution of $(3)$ too?
Clearly, we have $$AQ_i=Q_{i+1}H_i=Q_{i+1}U_iR_i\tag5$$ and hence $$AQ_i\lambda_i=\left\|r_0\right\|Q_{i+1}U_iU_i^\ast e_1\tag6\;,$$ Now, if I didn't made any mistake, I obtain $$(3)\Leftrightarrow\forall\mu\in\mathbb C^i:\langle r_0-AQ_i\lambda_i,\mu\rangle=0\tag7$$ (where I've used the result of my other question).
However, I don't know how I need to proceed.