Numerical recipes

11.2 Reduction of a Symmetric Matrix to Tridiagonal Form 471
where e1 is the unit vector [1, 0,...,0] T , and the choice of signs will be made
later. Then
P · x = x − u
H · (x ∓|x|e1) T · x
= x − 2u · (|x|2 ∓|x|x1)
2|x| 2 ∓ 2|x|x1
= x − u
= ±|x|e1
(11.2.6)
This shows that the Householder matrix P acts on a given vector x to zero all its
elements except the first one.
To reduce a symmetric matrix A to tridiagonal form, we choose the vector x
for the first Householder matrix to be the lower n − 1 elements of the first column.
Then the lower n − 2 elements will be zeroed:
⎡
1
⎢ 0
⎢
P1 · A = ⎢ 0
⎢
⎣ .
0 0 ···
(n−1)
P1
⎤ ⎡
0
⎥ ⎢
⎥ ⎢ a21
⎥ ⎢
⎥ · ⎢ a31 ⎢
⎥ ⎢
⎦ ⎣ . irrelevant
0
an1
=
⎡
a11
⎢ k
⎢ 0
⎢
⎣ .
0
a12 a13 ···
irrelevant
⎤
a1n
⎥
⎦
a11 a12 a13 ··· a1n
⎤
⎥
⎦
(11.2.7)
Here we have written the matrices in partitioned form, with (n−1) P denoting a
Householder matrix with dimensions (n − 1) × (n − 1). The quantity k is simply
plus or minus the magnitude of the vector [a 21,...,an1] T .
The complete orthogonal transformation is now
A ′ ⎡
⎤
a11
⎢
k
⎢
= P · A · P = ⎢ 0
⎢
⎣ .
0
k 0 ···
irrelevant
0
⎥
⎦
We have used the fact that P T = P.
(11.2.8)
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