- #1
mathmari
Gold Member
MHB
- 5,049
- 7
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:
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:
