It's a task i do to understand minimizing the error including the QR transformation with the help of Householder transformation. I think i really do something wrong but i dont get it running i hope you can help me.
Given the task to minimizing the error of
$$ A = \begin{pmatrix}
1 &1\\
1 & 0\\
1 & 1 \\
1 & 4
\end{pmatrix}
b=\left[2,\frac{1}{2},\frac{3}{2},5\right]^T
$$
So at first getting $A^TA = \begin{pmatrix} 4 & 6 \\ 6 &18 \end{pmatrix}$
after this calculate the householder vector $v_1=a_1+\alpha*e_1$.
$\alpha = sign(a_{1,1})*\lVert a_{1}\rVert = \sqrt{52} $
so
$$ v_1 = \begin{pmatrix} 4 \\ 6 \end{pmatrix} + \sqrt{52}*\begin{pmatrix}1 \\ 0 \end{pmatrix}$$
BUT the problem is if i calculate Q ouf of this:
$$Q = I- \frac{2*v*v^T}{v^T*v} $$
I would be done. It would be $A^TA = QR$ with $ R= QA^TA$. (Sure i am resulting in some matix like $R=\begin{pmatrix} x &y\\0&z\end{pmatrix}$
But if i calculate that with Matlab i get incorrect results ($R=A^Tb$). Even the calculation of $v$ seems to be incorrect since it should ne just simple in fact that no calculater is allowed. If i simply calculate $A^TAx=A^Tb$ it's a simple calculation resulting in $x_1=\frac{7}{12} x_2=\frac{10}{9}$ which i dont get with the QR in any way.
Best Answer
Based on the comments, this is probably what you're supposed to do.
To transform $A$ to the upper triangular form requires an application of two Householder transformations. Take $$ Q_1=I-2\frac{v_1v_1^T}{v_1^Tv_1}= \frac{1}{6} \begin{bmatrix} -3 & -3 & -3 & -3 \\ -3 & 5 & -1 & -1 \\ -3 & -1 & 5 & -1 \\ -3 & -1 & -1 & 5 \\ \end{bmatrix} \quad\text{with}\quad v_1=[3,1,1,1]^T. $$ This gives $$ Q_1^TA=\begin{bmatrix} -2 & -3 \\ 0 & -4/3 \\ 0 & -1/3 \\ 0 & 8/3 \end{bmatrix}. $$ Next, take $$ Q_2=I-2\frac{v_2v_2^T}{v_2^Tv_2} =\frac{1}{117}\begin{bmatrix} 117 & 0 & 0 & 0 \\ 0 & -52 & -13 & 104 \\ 0 & -13 & 116 & 8 \\ 0 & 104 & 8 & 53 \end{bmatrix} \quad\text{with}\quad v_2=[0,-13/3,-1/3,8/3]^T $$ to get $$ R:=Q^TA:=Q_2^TQ_1^TA= \begin{bmatrix} -2 & -3 \\ 0 & 3 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}. $$ Apply $Q^T$ to $b$ to get $$ Q^Tb= \begin{bmatrix} -9/2 \\ 10/3 \\ -11/39 \\ -19/78 \end{bmatrix}. $$ Hence, the least squares problem $Ax=b$ is equivalent to the least squares problem $$ \begin{bmatrix} -2 & -3 \\ 0 & 3 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}x = \begin{bmatrix} -9/2 \\ 10/3 \\ -11/39 \\ -19/78 \end{bmatrix}. $$ The residual 2-norm is minimal, if the first two components of the residual vector are zero, that is, the least squares solution is given by the solution of $$ \begin{bmatrix} -2 & -3 \\ 0 & 3 \end{bmatrix}x = \begin{bmatrix} -9/2 \\ 10/3 \end{bmatrix}. $$