Introduction to Computational Linear Algebra

Orthogonal Factorizations and Linear Least Squares Problems 89

where for x ∈ R, sign(x) = 1 for x ≥ 0 and sign(x) = −1 for x < 0. Thus:

v = w + sign(w

1

)kwke

1

, (4.28)

and from (4.26) one obtains:

Hw = −sign(w

1

)kwke

1

, (4.29)

Thus, H annihilates all the components of w, except the first one. If:

w =









w

1

w

2

...

w

n









,

then:

Hw = w − 2

v

T

w

v

T

v

v =









−sign(w

1

)kwk

0

...

0









.

On the other hand, the matrix H satisfies the following properties:

Proposition 4.1 The matrix H = I −

2

v

T

v

vv

T

is symmetric and orthogonal.

Therefore H

T

= H

−1

= H and H

2

= I. Every vector of the hyperplane

(v)

⊥

is invariant by H and Hv = −v: H is the orthogonal reflector w.r.t. the

hyperplane (v)

⊥

.

Pro ...