Linear Algebra, Theory And Applications, 2012a

15.2. THE QR ALGORITHM 389
Proof: Let
Q =(q 1 , ··· , q n ) ,Q (k) = ( q k 1, ··· , q k )
n
where the q are the columns. Also denote by rij k the ijth entry of R k . Thus
⎛
Q (k) R k = ( q k 1, ··· , q k ) ⎜
n ⎝
r k 11
. ..
∗
0 r k nn
⎞
⎟
⎠
It follows
and so
Therefore,
Next consider the second column.
r k 11q k 1 → q 1
r11 k = ∣ r
k
11 q k ∣
1 → 1
q k 1 → q 1 .
r k 12q k 1 + r k 22q k 2 → q 2
Taking the inner product of both sides with q k 1 it follows
(
lim
k→∞ rk 12 = lim q2 · q k )
1 =(q2 · q 1 )=0.
k→∞
Therefore,
lim
k→∞ rk 22q k 2 = q 2
and since r k 22 > 0, it follows as in the first part that r k 22 → 1. Hence
Continuing this way, it follows
for all i ≠ j and
lim
k→∞ rk jj =1,
lim
k→∞ qk 2 = q 2 .
lim
k→∞ rk ij =0
lim
k→∞ qk j = q j .
Thus R k → I and Q (k) → Q. This proves the first part of the lemma.
The second part follows immediately. If QR = Q ′ R ′ = A where A −1 exists, then
Q ∗ Q ′ = R (R ′ ) −1
and I need to show both sides of the above are equal to I. The left side of the above is
unitary and the right side is upper triangular having positive entries on the diagonal. This
is because the inverse of such an upper triangular matrix having positive entries on the
main diagonal is still upper triangular having positive entries on the main diagonal and
the product of two such upper triangular matrices gives another of the same form having
positive entries on the main diagonal. Suppose then that Q = R where Q is unitary and R
is upper triangular having positive entries on the main diagonal. Let Q k = Q and R k = R.
It follows
IR k → R = Q
and so from the first part, R k → I but R k = R and so R = I. Thus applying this to
Q ∗ Q ′ = R (R ′ ) −1 yields both sides equal I.