48.
Jacobi’s Method for Diagonalizing a Symmetric Matrix
Let A be symmetric and apq the largest (in absolute value) off-diagonal entry.
With A0 = A, we want to find a matrix U of the form
p q i j
p cos θ sin θ 0 0
U= q − sin θ cos θ 0 0
i 0 0 1 0
j 0 0 0 1
such that the matrix
A1 = UT AU
has a1pq = 0. Note that UT U = I and A1 is symmetric.
Choose θ such that − π4 < θ ≤ π4 and
2 sin θ cos θ 2apq
tan 2θ = = .
cos2 θ − sin2 θ aqq − app
Then, using the abbreviations c = cos θ and s = sin θ, we have A1 =
p q i j
p c2 app − 2csapq + s2 aqq cs(app − aqq ) capi capj
+(c2 − s2 )apq −saqi −saqj
q cs(app − aqq ) + (c2 − s2 )apq s2 app + 2csapq sapi + caqi sapj
+c2 aqq +caqj
i caip − saiq saip + caiq aii aij
j cajp − sajq sajp + cajq aji ajj
By the choice of θ we have a1pq = 0. We may iterate the process, forming a sequence
of matrices A0 , A1 , A2 , . . . until the off diagonal entries are sufficiently small. The
limit of this sequence is a diagonal matrix D = diag(λ1 , . . . , λn ).
49. Householder Reduction
For larger symmetric matrices, say bigger than 10 × 10, there is a more efficient
method for finding the eigenvalues. This method is somewhat more complicated
than the Jacobi algorithm, but it is widely used. It is based on the following result.
1
2
Householder Reduction Theorem. For any square matrix A there exists an
orthogonal matrix U such that B = UT AU has bij = 0 for j ≤ i − 2, something
like
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
UT AU = 0 ∗ ∗ ∗ ∗
0 0 ∗ ∗ ∗
0 0 0 ∗ ∗
This is called Hessenberg form. If A is symmetric, then B is tridiagonal and
symmetric.
Outline of proof. We will define n − 2 orthogonal matrices P1 , . . . , Pn−2 . Each of
these will have the special form
Pi = I − 2wi wiT
for a suitably chosen vector wi of length 1. Let
A1 = A
A2 = P1 A1 PT1
..
.
An−1 = Pn−2 An−2 PTn−2 .
Since PTi = P−1
i , these matrices are all similar and hence have the same eigenvalues.
Moreover,
An−1 = Pn−2 . . . P1 APT1 . . . PTn−2
= UAUT .
If we choose the wi ’s correctly, we can get B = An−1 to have zeroes below the
subdiagonal. ¤
If A is symmetric, then after the Householder reduction we have the tridiagonal
matrix
a1 b1 0
b1 a2 b2
.. .. ..
An−1 = . . . .
bn−2 an−1 bn−1
0 bn−1 an