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Linear Algebra, Theory And Applications, 2012a

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15.2. THE QR ALGORITHM 393<br />

Corollary 15.2.5 Let A be a real symmetric n × n matrix having eigenvalues<br />

λ 1 >λ 2 > ···>λ n > 0<br />

and let Q be defined by<br />

QDQ T = A, D = Q T AQ, (15.14)<br />

where Q is orthogonal and D is a diagonal matrix having the eigenvalues on the main<br />

diagonal decreasing in size from the upper left corner to the lower right. Let Q T have an<br />

LU factorization. Then in the QR algorithm, the matrices Q (k) converge to Q ′ where Q ′ is<br />

thesameasQ except having some columns multiplied by (−1) . Thus the columns of Q ′ are<br />

eigenvectors of A. The matrices A k converge to D.<br />

Proof: This follows from Theorem 15.2.4. Here S = Q, S −1 = Q T . Thus<br />

Q = S = QR<br />

and R = I. By Theorem 15.2.4 and Lemma 15.2.2,<br />

A k = Q (k)T AQ (k) → Q ′T AQ ′ = Q T AQ = D.<br />

because formula (15.14) is unaffected by replacing Q with Q ′ . <br />

When using the QR algorithm, it is not necessary to check technical condition about<br />

S −1 having an LU factorization. The algorithm delivers a sequence of matrices which are<br />

similar to the original one. If that sequence converges to an upper triangular matrix, then<br />

the algorithm worked. Furthermore, the technical condition is sufficient but not necessary.<br />

The algorithm will work even without the technical condition.<br />

Example 15.2.6 Find the eigenvalues and eigenvectors of the matrix<br />

⎛<br />

A = ⎝ 5 1 1 ⎞<br />

1 3 2 ⎠<br />

1 2 1<br />

It is a symmetric matrix but other than that, I just pulled it out of the air. By Lemma<br />

15.2.2 it follows A k = Q (k)T AQ (k) . <strong>And</strong> so to get to the answer quickly I could have the<br />

computer raise A toapowerandthentaketheQR factorization of what results to get the<br />

k th iteration using the above formula. Lets pick k =10.<br />

⎛<br />

⎝ 5 1 1 3 1<br />

2<br />

⎞<br />

⎠<br />

1 2 1<br />

10<br />

⎛<br />

= ⎝ 4. 227 3 × ⎞<br />

107 2. 595 9 × 10 7 1. 861 1 × 10 7<br />

2. 595 9 × 10 7 1. 607 2 × 10 7 1. 150 6 × 10 7 ⎠<br />

1. 861 1 × 10 7 1. 150 6 × 10 7 8. 239 6 × 10 6<br />

Now take QR factorization of this. The computer will do that also.<br />

This yields<br />

⎛<br />

⎞<br />

. 797 85 −. 599 12 −6. 694 3 × 10−2<br />

⎝ . 489 95 . 709 12 −. 507 06 ⎠ ·<br />

. 351 26 . 371 76 . 859 31<br />

⎛<br />

⎝ 5. 298 3 × ⎞<br />

107 3. 262 7 × 10 7 2. 338 × 10 7<br />

0 1. 217 2 × 10 5 71946. ⎠<br />

0 0 277. 03

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