MHB Householder method : Which α do we take?

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The discussion focuses on the Householder method for QR factorization of the matrix A. The first column is analyzed to determine the value of α, where it is concluded that α should be -3 to maximize the norm difference. For the second iteration, when considering the submatrix, both possible values of α yield equal norms, leading to uncertainty about which to choose. It is clarified that while different choices may result in distinct QR decompositions, the impact on calculation errors remains minimized. Ultimately, the choice of α in this scenario does not affect the error significantly.
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Hey! 😊

We have the matrix $A=\begin{pmatrix}1 & 4 & -1 \\ 2 & 2 & 7 \\ 2 & -4 & 7\end{pmatrix}$ and we want to calculate the $QR$ factorisation using the Householder method.

First we take the first column of the matrix $a_0=\begin{pmatrix}1 \\ 2 \\ 2\end{pmatrix}$.
We have that $\alpha=\pm \|a_0\|=\pm \sqrt{9}\pm 3$.
Then we take the sign of that so that $\|a_0-\alpha e_1\|_2$ is big.
We have $$\|a_0-\alpha e_1\|_2=\begin{cases}\|\begin{pmatrix}1 \\ 2 \\ 2\end{pmatrix}-\begin{pmatrix}3 \\ 0 \\ 0\end{pmatrix}\|_2 =\|\begin{pmatrix}-2 \\ 2 \\ 2\end{pmatrix} =\sqrt{12}\\ \|\begin{pmatrix}1 \\ 2 \\ 2\end{pmatrix}-\begin{pmatrix}-3 \\ 0 \\ 0\end{pmatrix}\|_2 =\|\begin{pmatrix}4 \\ 2 \\ 2\end{pmatrix}\end{cases} =\sqrt{20}$$
So we take $\alpha=-3$.
Then $u=\frac{1}{\|a_0-\alpha e_1\|_2}\begin{pmatrix}4 \\ 2 \\ 2\end{pmatrix}=\begin{pmatrix}\frac{4}{\sqrt{20}} \\ \frac{2}{\sqrt{20}} \\ \frac{2}{\sqrt{20}}\end{pmatrix}$.
Then $$H_1=I-2uu^T=\cdots =\begin{pmatrix}-1/3 & -2/3 & -2/3 \\ -2/3 & 2/3 & -1/3 \\ -2/3 & -1/3 & 2/3\end{pmatrix}$$
Then $A^{(1)}=H_1\cdot A=\begin{pmatrix}-3 & 0 & -9 \\ 0 & 0 & 3 \\ 0 & -6 & 3\end{pmatrix}$.
Then we consider the first column of the $2\times 2$-submatrix, $a_1=\begin{pmatrix}0 \\ -6\end{pmatrix}$.
Then $\alpha=\pm \|a_1\|=\pm 6$.
We have $$\|a_1-\alpha e_1\|_2=\begin{cases}\|\begin{pmatrix}0 \\ -6\end{pmatrix}-\begin{pmatrix}6\\ 0\end{pmatrix}\|_2 =\|\begin{pmatrix}-6 \\ -6\end{pmatrix} =6\sqrt{2}\\ \|\begin{pmatrix}0\\ -6\end{pmatrix}-\begin{pmatrix}-6 \\ 0\end{pmatrix}\|_2 =\|\begin{pmatrix}6 \\ 0\end{pmatrix}\end{cases} =6\sqrt{2}$$
In this case the norms are equal, which $\alpha$ do we take then?

:unsure:
 
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mathmari said:
In this case the norms are equal, which $\alpha$ do we take then?
Since the norms are equal, the impact on the calculation error is the same.
So it doesn't matter which one we pick. You can choose. 🤔
 
Klaas van Aarsen said:
Since the norms are equal, the impact on the calculation error is the same.
So it doesn't matter which one we pick. You can choose. 🤔

But the result that we get will be different, or not? :unsure:
 
mathmari said:
But the result that we get will be different, or not?
We may find a different QR decomposition, but the impact of calculation errors will have been minimized to the same extent. 🤔
 
Klaas van Aarsen said:
We may find a different QR decomposition, but the impact of calculation errors will have been minimized to the same extent. 🤔

Ok! Thank you! 🤩
 
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