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# ordqz

Reorder eigenvalues in QZ factorization

## Description

[AAS,BBS,QS,ZS] = ordqz(AA,BB,Q,Z,select) reorders the QZ factorization Q*A*Z = AA and Q*B*Z = BB produced by [AA,BB,Q,Z] = qz(A,B) and returns the reordered matrix pair (AAS,BBS) along with orthogonal matrices (QS,ZS), such that QS*A*ZS = AAS and QS*B*ZS = BBS.

In this reordering, the selected cluster of eigenvalues appears in the leading (upper left) diagonal blocks of the quasitriangular pair (AAS,BBS). The leading columns of ZS span the corresponding invariant subspace. The logical vector select specifies the selected cluster as e(select), where e = ordeig(AA,BB).

example

[AAS,BBS,QS,ZS] = ordqz(AA,BB,Q,Z,keyword) sets the selected cluster to include all eigenvalues in the region specified by keyword.

[AAS,BBS,QS,ZS] = ordqz(AA,BB,Q,Z,clusters) reorders multiple clusters simultaneously. ordqz sorts the specified clusters in descending order along the diagonal of (AAS,BBS), with the cluster of highest index appearing in the upper left corner.

## Examples

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Compute the QZ factorization of a pair of matrices, and then reorder the factors according to a specified ordering of the eigenvalues.

Find the QZ factorization, or generalized Schur decomposition, of a pair of matrices A and B. This decomposition results in the factors $\mathrm{AA}=\mathrm{QAZ}$ and $\mathrm{BB}=\mathrm{QBZ}$.

A = magic(5);
B = hilb(5);
[AA,BB,Q,Z] = qz(A,B)
AA = 5×5

14.5272   -2.3517    8.5757   -0.2350   -1.4432
0  -19.7471    2.1824    4.5417    7.2059
0         0  -17.9538    8.9292   -9.6961
0         0         0   30.3449  -47.9191
0         0         0         0   32.4399

BB = 5×5

0.0000    0.0005    0.0018    0.0465    0.2304
0    0.0008    0.0199    0.1662    0.7320
0         0    0.0210    0.1006   -0.1341
0         0         0    0.0623   -1.1380
0         0         0         0    0.7434

Q = 5×5

-0.1743   -0.1099   -0.0789   -0.4690    0.8552
-0.7567   -0.1151   -0.0846    0.6172    0.1617
-0.4010    0.6782    0.5478   -0.2664   -0.0901
0.4178   -0.0297    0.6473    0.4883    0.4089
-0.2484   -0.7168    0.5173   -0.2995   -0.2593

Z = 5×5

0.0057   -0.0424   -0.2914   -0.5860   -0.7549
-0.1125    0.4109    0.7635    0.1734   -0.4533
0.4995   -0.6746    0.1486    0.4053   -0.3303
-0.7694   -0.2140   -0.2614    0.4749   -0.2616
0.3818    0.5731   -0.4917    0.4866   -0.2173

Since AA and BB are triangular, use ordeig to extract the eigenvalues from the diagonal blocks of AA and BB.

e = ordeig(AA,BB)
e = 5×1
106 ×

2.8871
-0.0257
-0.0009
0.0005
0.0000

Separate the eigenvalues into clusters, with the real positive eigenvalues ($\mathit{e}>0$) forming the leading cluster. Reorder the matrices AA, BB, Q, and Z according to this ordering of the eigenvalues.

[AAS,BBS,QS,ZS] = ordqz(AA,BB,Q,Z,'rhp')
AAS = 5×5

14.5272    1.2849    1.0391    7.6821    4.4119
0   21.7128   19.1784   -1.8380   -9.1187
0         0   60.3083   -8.4452   -6.4304
0         0         0  -18.2081   -3.3783
0         0         0         0  -14.6375

BBS = 5×5

0.0000   -0.0114    0.1908   -0.1119    0.0788
0    0.0446   -0.0377    0.1107   -0.1978
0         0    1.3820   -0.6325    0.2807
0         0         0    0.0007    0.0137
0         0         0         0    0.0171

QS = 5×5

-0.1743   -0.1099   -0.0789   -0.4690    0.8552
0.6353   -0.1853   -0.4099   -0.5765   -0.2483
-0.7034   -0.4518   -0.3456   -0.2295   -0.3591
-0.1415    0.2036    0.7054   -0.6065   -0.2703
-0.2263    0.8414   -0.4568   -0.1647   -0.0705

ZS = 5×5

0.0057    0.0088   -0.5288    0.3591   -0.7690
-0.1125    0.6095   -0.3858    0.4737    0.4926
0.4995   -0.6478   -0.2711    0.3644    0.3529
-0.7694   -0.4176   -0.4090   -0.1750    0.1890
0.3818    0.1855   -0.5752   -0.6952    0.0758

Examine the new eigenvalue order.

E2 = ordeig(AAS,BBS)
E2 = 5×1
106 ×

2.8871
0.0005
0.0000
-0.0257
-0.0009

## Input Arguments

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Matrix factors, specified as the matrices returned by [AA,BB,Q,Z] = qz(A,B). These matrices satisfy Q*A*Z = AA and Q*B*Z = BB. For complex matrices, AA and BB are triangular.

If AA and BB do not form a valid QZ decomposition, then ordqz does not produce an error and returns incorrect results.

Data Types: single | double
Complex Number Support: Yes

Unitary matrices, specified as the matrices returned by [AA,BB,Q,Z] = qz(A,B). These matrices satisfy Q*A*Z = AA and Q*B*Z = BB.

Data Types: single | double
Complex Number Support: Yes

Cluster selector, specified as a logical vector with length equal to the number of generalized eigenvalues. The generalized eigenvalues appear along the diagonal of AA-λ*BB.

Data Types: logical

Eigenvalue region keyword, specified as one of the options in this table.

Option

Selected Region

(e = ordeig(AA,BB))

'lhp'

Left-half plane (real(e) < 0)

'rhp'

Right-half plane (real(e) >= 0)

'udi'

Interior of unit disk (abs(e) < 1)

'udo'

Exterior of unit disk (abs(e) >= 1)

Cluster indices, specified as a vector of positive integers with length equal to the number of eigenvalues. clusters assigns each eigenvalue returned by e = ordeig(AA,BB) to a different cluster. All eigenvalues with the same index value in clusters form one cluster.

Example: ordqz(AA,BB,Q,Z,[1 1 2 3 3]) groups five eigenvalues into three clusters.

Data Types: single | double

## Output Arguments

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Reordered matrices, returned as matrices that satisfy QS*A*ZS = AAS and QS*B*ZS = BBS.

QS and ZS are unitary, while AAS is quasitriangular and BBS is triangular.

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### Quasitriangular

An upper quasitriangular matrix can result from the Schur decomposition or generalized Schur (QZ) decomposition of real matrices. These matrices are block upper triangular, with 1-by-1 and 2-by-2 blocks along the diagonal.

The eigenvalues of these diagonal blocks are also the eigenvalues of the matrix. The 1-by-1 blocks correspond to real eigenvalues, and the 2-by-2 blocks correspond to complex conjugate eigenvalue pairs.

## Tips

• If AA has complex conjugate pairs (nonzero elements on the subdiagonal), then you should move the pair to the same cluster. Otherwise, ordqz acts to keep the pair together:

• If select is not the same for two eigenvalues in a conjugate pair, then ordqz treats both as selected.

• If clusters is not the same for two eigenvalues in a conjugate pair, then ordqz treats both as part of the cluster with larger index.