Find a matrix by Householder transformation

Question

Let A = \begin{bmatrix} 1 & -2 \\0 & 0.84 \\0 & 2.88 \\0 & 0 \\\end{bmatrix}

find matrix B that

B * A = \begin{bmatrix} * & * \\0 & * \\0 & 0 \\0 & 0 \\\end{bmatrix}

Could you help me out with that? I would really appreaciate it. It is supposed to be a simple task but I fail every time I try to do it.

Answer 1

Here is a solution :

$$B=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0.28 & \ \ \ 0.96 & 0\\ 0 & 0.96 & -0.28 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \ \text{giving} \ \ B*A=\begin{pmatrix} 1 & -2\\ 0 & \ \ 3\\ 0 & \ \ 0\\ 0 & \ \ 0 \end{pmatrix}. $$

I assume a certain knowledge about Householder transforms, in particular that such transforms are isometries, thus are norm-preserving ; therefore we must care to send vectors onto vectors with the same norm.

In order for you to understand how $B$ has been found but also to leave you some work :)), here is the Matlab program that has helped me to do the computations:

#1:  U=[1;0;0;0];V=[-2;0.84;2.88;0];% first and second col. of matrix A
#2:  A=[U,V];
#3:  a=sqrt(V(2)^2+V(3)^2+V(4)^2);% observe that V(1)^2 is "missing"
#4:  V1=[-2;a;0;0]; % norm(V1)=norm(V) (Householder = isometric transf.)
#5:  V2=V-V1;
#6:  V2=V2/norm(V2);% now with unit norm
#7:  I4=eye(4); % order 4 identity matrix
#8:  B=I4-2*V2*V2';% Householder matrix mapping V onto V1 and U onto itself.

Important remark : (in connection with the remark made by @Dunham) Let us recall that the $k$th step of the Householder method consists in working on an $k-1$ dimensional space (here, $k=4$: we are in a first step) This is why (program line #3) we haven't modified the first coordinate of the second vector V (said otherwise: the first coordinate of $V_2:=V-V_1$ is $0$, thus $U$ is sent onto itself because $(I_-V_2V_2^T)U=IU-0=U$).

Remark: Results are exact (no accuracy loss) because in fact all computations can be done with rational numbers with common denominator $25$ such as:

$$ 0.84=\tfrac{21}{25}, \ \ 2.88=\tfrac{72}{25}, \ \ 0.24=\tfrac{7}{25} \ \ \text{and} \ \ 0.96=\tfrac{24}{25}$$