THESE de DOCTORAT - cerfacs
THESE de DOCTORAT - cerfacs
THESE de DOCTORAT - cerfacs
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3.3 Solving the system Ax = b 49<br />
where ̂Q m+1 ∈ C n×m+1 is an orthonormal matrix and ˜H m ∈ C m+1×m is an upper-Hessenberg<br />
matrix 4 .<br />
̂D y ˜H y<br />
̂Q m+1<br />
m<br />
m +1× m<br />
m<br />
n × m<br />
n × m +1<br />
Figure 3.6: Hessenberg <strong>de</strong>composition<br />
The least square problem is now given by<br />
Find y such that ‖r 0 − Q m+1 ˜H m y‖ 2 = ‖r m ‖ 2 is minimized (3.21)<br />
Let us introduce r 0 = q 1 ‖r 0 ‖ (see Eq. 3.43). The vector q 1 can be expressed as Q m+1 e 1 = q 1<br />
where e 1 is the canonical unit vector e 1 = (1, 0, 0, ...). Furthermore, the invariance of inner<br />
products means that the angles between vectors are preserved, and so are their lengths:<br />
‖Qx‖ = ‖x‖. The final expression results in<br />
‖Q m+1<br />
(<br />
‖r0 ‖e 1 − ˜H m y ) ‖ 2 = ‖ ( ‖r‖e 1 − ˜H m y ) ‖ 2 (3.22)<br />
leading to<br />
Find y such that ‖ ( ‖r‖e 1 − ˜H m y ) ‖ 2 = ‖r m ‖ 2 is minimized (3.23)<br />
Once y is found through the least square methodology, a solution for the Ax m ≈ b problem can<br />
be found by solving x m = x 0 + ̂Q m y. There are still two crucial issues to resolve:<br />
1. How to construct such matrices ̂Q m and ˜H m ?<br />
2. What is the ‘least square problem’ about ?<br />
These two issues will be addressed in sections (3.3.1) and (3.3.2) respectively.<br />
4 An upper- Hessenberg matrix H is a matrix with zeros below the first subdiagonal