Permuted max-eigenvector problem is NP -complete P.Butkovi c - PowerPoint PPT Presentation

MMIPP: Steady state The system is in a steady state if it is moving forward in regular steps Equivalently, if there is a λ such that x ( r + 1 ) = λ � x ( r ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

MMIPP: Steady state The system is in a steady state if it is moving forward in regular steps Equivalently, if there is a λ such that x ( r + 1 ) = λ � x ( r ) Since x ( r + 1 ) = A � x ( r ) ( r = 0 , 1 , . . . ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

MMIPP: Steady state The system is in a steady state if it is moving forward in regular steps Equivalently, if there is a λ such that x ( r + 1 ) = λ � x ( r ) Since x ( r + 1 ) = A � x ( r ) ( r = 0 , 1 , . . . ) x ( 0 ) should satisfy A � x = λ � x P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

MMIPP: Steady state The system is in a steady state if it is moving forward in regular steps Equivalently, if there is a λ such that x ( r + 1 ) = λ � x ( r ) Since x ( r + 1 ) = A � x ( r ) ( r = 0 , 1 , . . . ) x ( 0 ) should satisfy A � x = λ � x One-o¤ process ( b is the vector of completion times ): A � x = b P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Two basic problems Problem (LINEAR SYSTEM [LS]) m …nd all x 2 R n satisfying m � n and b 2 R Given A 2 R A � x = b P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Two basic problems Problem (LINEAR SYSTEM [LS]) m …nd all x 2 R n satisfying m � n and b 2 R Given A 2 R A � x = b Problem (EIGENVECTOR [EV]) n � n …nd all x 2 R n , x 6 = ( ε , ..., ε ) T (eigenvectors) Given A 2 R such that A � x = λ � x for some λ 2 R (eigenvalue) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Permuted matrices and vectors For n � n A = ( a ij ) 2 R ( x 1 , ..., x n ) T 2 R n x = 2 π , σ P n de…ne � � A ( π , σ ) = a π ( i ) , σ ( j ) � � T x ( π ) = x π ( 1 ) , ..., x π ( n ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Permuted basic problems Problem (PERMUTED LINEAR SYSTEM [PLS]) m � n and b 2 R m , is there a π 2 P m such that Given A 2 R A � x = b ( π ) has a solution? P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Permuted basic problems Problem (PERMUTED LINEAR SYSTEM [PLS]) m � n and b 2 R m , is there a π 2 P m such that Given A 2 R A � x = b ( π ) has a solution? Problem (PERMUTED EIGENVECTOR [PEV]) n � n and x 2 R n , x 6 = ( ε , ..., ε ) T , is there a π 2 P n Given A 2 R such that A � x ( π ) = λ � x ( π ) for some λ 2 R ? P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Permuted basic problems - integer versions Problem (INTEGER PERMUTED LINEAR SYSTEM [IPLS]) Given A 2 Z m � n and b 2 Z m , is there a π 2 P m such that A � x = b ( π ) has a solution ? P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Permuted basic problems - integer versions Problem (INTEGER PERMUTED LINEAR SYSTEM [IPLS]) Given A 2 Z m � n and b 2 Z m , is there a π 2 P m such that A � x = b ( π ) has a solution ? Problem (INTEGER PERMUTED EIGENVECTOR [IPEV]) Given A 2 Z n � n and x 2 Z n , is there a π 2 P n such that A � x ( π ) = x ( π ) ? P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Permuted basic problems - integer versions Problem (INTEGER PERMUTED LINEAR SYSTEM [IPLS]) Given A 2 Z m � n and b 2 Z m , is there a π 2 P m such that A � x = b ( π ) has a solution ? Problem (INTEGER PERMUTED EIGENVECTOR [IPEV]) Given A 2 Z n � n and x 2 Z n , is there a π 2 P n such that A � x ( π ) = x ( π ) ? Theorem Both IPEV and IPLS are NP-complete. P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

BANDWIDTH Problem (BANDWIDTH) Given an undirected graph G = ( N , E ) and a positive integer K � n , is there a π 2 P n such that j π ( u ) � π ( v ) j � K for all uv 2 E ? Equivalently: P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

BANDWIDTH Problem (BANDWIDTH) Given an undirected graph G = ( N , E ) and a positive integer K � n , is there a π 2 P n such that j π ( u ) � π ( v ) j � K for all uv 2 E ? Equivalently: Problem (BANDWIDTH - MATRIX VERSION) Given an n � n symmetric 0 � 1 matrix M = ( m ij ) with zero diagonal, and a positive integer K � n , is there a π 2 P n such that j i � j j � K whenever m π ( i ) , π ( j ) = 1 ? P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

BANDWIDTH 0 K 0 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of linear systems Let A 2 R m � n and b 2 R m , M = f 1 , ..., m g , N = f 1 , ..., n g A � x = b P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of linear systems Let A 2 R m � n and b 2 R m , M = f 1 , ..., m g , N = f 1 , ..., n g A � x = b S ( A , b ) = f x 2 R n ; A � x = b g P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of linear systems Let A 2 R m � n and b 2 R m , M = f 1 , ..., m g , N = f 1 , ..., n g A � x = b S ( A , b ) = f x 2 R n ; A � x = b g x j = � max ( a ij � b i ) , j 2 N i P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of linear systems Let A 2 R m � n and b 2 R m , M = f 1 , ..., m g , N = f 1 , ..., n g A � x = b S ( A , b ) = f x 2 R n ; A � x = b g x j = � max ( a ij � b i ) , j 2 N i x = ( x 1 , ..., x n ) T P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of linear systems Let A 2 R m � n and b 2 R m , M = f 1 , ..., m g , N = f 1 , ..., n g A � x = b S ( A , b ) = f x 2 R n ; A � x = b g x j = � max ( a ij � b i ) , j 2 N i x = ( x 1 , ..., x n ) T M j = f k 2 M ; a kj � b k = max ( a ij � b i ) g , j 2 N i P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of max-linear systems Theorem (R.A.Cuninghame-Green) Let A 2 R m � n , b 2 R m and x 2 R n . Then x 2 S ( A , b ) if and only if (a) x � x and (b) [ M j = M x j = x j Corollary (1) The following three statements are equivalent: P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of max-linear systems Theorem (R.A.Cuninghame-Green) Let A 2 R m � n , b 2 R m and x 2 R n . Then x 2 S ( A , b ) if and only if (a) x � x and (b) [ M j = M x j = x j Corollary (1) The following three statements are equivalent: (I) S ( A , b ) 6 = ∅ P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of max-linear systems Theorem (R.A.Cuninghame-Green) Let A 2 R m � n , b 2 R m and x 2 R n . Then x 2 S ( A , b ) if and only if (a) x � x and (b) [ M j = M x j = x j Corollary (1) The following three statements are equivalent: (I) S ( A , b ) 6 = ∅ (II) x 2 S ( A , b ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Solvability of max-linear systems Theorem (R.A.Cuninghame-Green) Let A 2 R m � n , b 2 R m and x 2 R n . Then x 2 S ( A , b ) if and only if (a) x � x and (b) [ M j = M x j = x j Corollary (1) The following three statements are equivalent: (I) S ( A , b ) 6 = ∅ (II) x 2 S ( A , b ) (III) S j 2 N M j = M P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Unique solution to a max-linear system Corollary (2) S ( A , b ) = f x g if and only if P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Unique solution to a max-linear system Corollary (2) S ( A , b ) = f x g if and only if (a) S j 2 N M j = M and P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Unique solution to a max-linear system Corollary (2) S ( A , b ) = f x g if and only if (a) S j 2 N M j = M and (b) S j 2 N 0 M j 6 = M for any N 0 � N , N 0 6 = N . P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Unique solution to a max-linear system Corollary (2) S ( A , b ) = f x g if and only if (a) S j 2 N M j = M and (b) S j 2 N 0 M j 6 = M for any N 0 � N , N 0 6 = N . Corollary (3) If m = n then S ( A , b ) = f x g if and only if there is a π 2 P n such that M π ( j ) = f j g for all j 2 N . Equivalently a i , π ( j ) � b i < a j , π ( j ) � b j for all i , j 2 N , i 6 = j . P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity A = ( a ij ) 2 R n � n is strongly regular i¤ ( 9 b 2 R n ) j S ( A , b ) j = 1 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity A = ( a ij ) 2 R n � n is strongly regular i¤ ( 9 b 2 R n ) j S ( A , b ) j = 1 Linear assignment problem for A : Find a π 2 P n maximising w ( A , π ) = ∑ a i , π ( i ) i 2 N P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity A = ( a ij ) 2 R n � n is strongly regular i¤ ( 9 b 2 R n ) j S ( A , b ) j = 1 Linear assignment problem for A : Find a π 2 P n maximising w ( A , π ) = ∑ a i , π ( i ) i 2 N ap ( A ) = f σ 2 P n ; w ( A , σ ) = max π 2 P n w ( A , π ) g P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity A = ( a ij ) 2 R n � n is strongly regular i¤ ( 9 b 2 R n ) j S ( A , b ) j = 1 Linear assignment problem for A : Find a π 2 P n maximising w ( A , π ) = ∑ a i , π ( i ) i 2 N ap ( A ) = f σ 2 P n ; w ( A , σ ) = max π 2 P n w ( A , π ) g (PB + Hevery) A = ( a ij ) 2 R n � n is strongly regular if and only if j ap ( A ) j = 1 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity A = ( a ij ) 2 R n � n is strongly regular i¤ ( 9 b 2 R n ) j S ( A , b ) j = 1 Linear assignment problem for A : Find a π 2 P n maximising w ( A , π ) = ∑ a i , π ( i ) i 2 N ap ( A ) = f σ 2 P n ; w ( A , σ ) = max π 2 P n w ( A , π ) g (PB + Hevery) A = ( a ij ) 2 R n � n is strongly regular if and only if j ap ( A ) j = 1 What are those b if A is strongly regular? P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity A = ( a ij ) 2 R n � n is strongly regular i¤ ( 9 b 2 R n ) j S ( A , b ) j = 1 Linear assignment problem for A : Find a π 2 P n maximising w ( A , π ) = ∑ a i , π ( i ) i 2 N ap ( A ) = f σ 2 P n ; w ( A , σ ) = max π 2 P n w ( A , π ) g (PB + Hevery) A = ( a ij ) 2 R n � n is strongly regular if and only if j ap ( A ) j = 1 What are those b if A is strongly regular? A 2 R n � n � ! S A = f b 2 R n ; A � x = b has a unique solution g P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g S A ... the simple image set (of the mapping x 7� ! A � x ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g S A ... the simple image set (of the mapping x 7� ! A � x ) A = ( a ij ) 2 R n � n is normalised i¤ P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g S A ... the simple image set (of the mapping x 7� ! A � x ) A = ( a ij ) 2 R n � n is normalised i¤ λ ( A ) = 0 ( A is de…nite) and P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g S A ... the simple image set (of the mapping x 7� ! A � x ) A = ( a ij ) 2 R n � n is normalised i¤ λ ( A ) = 0 ( A is de…nite) and a ii = 0 for all i 2 N ( A is increasing, A � I ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g S A ... the simple image set (of the mapping x 7� ! A � x ) A = ( a ij ) 2 R n � n is normalised i¤ λ ( A ) = 0 ( A is de…nite) and a ii = 0 for all i 2 N ( A is increasing, A � I ) ) ∆ ( A ) = Γ ( A ) = A � A 2 � ... � A n � 1 and A normalised = I � A � A 2 � ..., hence ∆ ( A ) = Γ ( A ) = A n � 1 = A n = A n + 1 = ... P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Strong regularity S A = f b 2 R n ; A � x = b has a unique solution g S A ... the simple image set (of the mapping x 7� ! A � x ) A = ( a ij ) 2 R n � n is normalised i¤ λ ( A ) = 0 ( A is de…nite) and a ii = 0 for all i 2 N ( A is increasing, A � I ) ) ∆ ( A ) = Γ ( A ) = A � A 2 � ... � A n � 1 and A normalised = I � A � A 2 � ..., hence ∆ ( A ) = Γ ( A ) = A n � 1 = A n = A n + 1 = ... A normalised = ) � A 2 � � Im � A 3 � � ... Im ( A ) � Im � A n � 1 � = Im ( A n ) = Im � A n + 1 � = ... = V ( A ) � Im P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set Theorem If A 2 R n � n is normalised and strongly regular (that is S A 6 = ∅ ) then V ( A ) = cl ( S A ) Corollary If A 2 R n � n is normalised and strongly regular then P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set Theorem If A 2 R n � n is normalised and strongly regular (that is S A 6 = ∅ ) then V ( A ) = cl ( S A ) Corollary If A 2 R n � n is normalised and strongly regular then A � b = b for every b 2 S A 1 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The simple image set Theorem If A 2 R n � n is normalised and strongly regular (that is S A 6 = ∅ ) then V ( A ) = cl ( S A ) Corollary If A 2 R n � n is normalised and strongly regular then A � b = b for every b 2 S A 1 For every b 2 V ( A ) there is a sequence f b ( k ) g ∞ k = 0 � S A such that 2 b ( k ) � ! b P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices � A , b ( k ) �� � � � b ( k ) 2 S A means � S � = 1 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices � A , b ( k ) �� � � � b ( k ) 2 S A means � S � = 1 If m = n : j S ( A , b ) j = 1 ( ) ( 9 π 2 P n ) M π ( j ) = f j g for all j 2 N P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices � A , b ( k ) �� � � � b ( k ) 2 S A means � S � = 1 If m = n : j S ( A , b ) j = 1 ( ) ( 9 π 2 P n ) M π ( j ) = f j g for all j 2 N Equivalently a i , π ( j ) � b ( k ) < a j , π ( j ) � b ( k ) i j for all i , j 2 N , i 6 = j . P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices � A , b ( k ) �� � � � b ( k ) 2 S A means � S � = 1 If m = n : j S ( A , b ) j = 1 ( ) ( 9 π 2 P n ) M π ( j ) = f j g for all j 2 N Equivalently a i , π ( j ) � b ( k ) < a j , π ( j ) � b ( k ) i j for all i , j 2 N , i 6 = j . If A is normalised and strongly regular then π = id , hence a ij � b ( k ) < � b ( k ) for every i , j 2 N , i 6 = j and k = 0 , 1 , ... . i j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices If A is normalised and strongly regular then a ij � b ( k ) < � b ( k ) for every i , j 2 N , i 6 = j and k = 0 , 1 , ... . i j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices If A is normalised and strongly regular then a ij � b ( k ) < � b ( k ) for every i , j 2 N , i 6 = j and k = 0 , 1 , ... . i j Let A = ( a ij ) 2 R n � n be normalised, strongly regular and b 2 R n . Then b 2 V ( A ) if and only if a ij � b i � � b j for every i , j 2 N P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

Normalised and strongly regular matrices If A is normalised and strongly regular then a ij � b ( k ) < � b ( k ) for every i , j 2 N , i 6 = j and k = 0 , 1 , ... . i j Let A = ( a ij ) 2 R n � n be normalised, strongly regular and b 2 R n . Then b 2 V ( A ) if and only if a ij � b i � � b j for every i , j 2 N Let A = ( a ij ) 2 Z n � n be normalised, strongly regular, b 2 Z n and π 2 P n . Then b ( π ) 2 V ( A ) if and only if a π ( i ) , π ( j ) � b i � b j for every i , j 2 N P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Theorem IPEV is NP-complete for the class of normalised, strongly regular matrices. Proof M = ( m ij ) , 0 < K � n ... an instance of BANDWIDTH. P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Theorem IPEV is NP-complete for the class of normalised, strongly regular matrices. Proof M = ( m ij ) , 0 < K � n ... an instance of BANDWIDTH. Let A = ( a ij ) 2 Z n � n : 8 � K if m ij = 1 < a ij = � n if m ij = 0 , i 6 = j : 0 if i = j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Theorem IPEV is NP-complete for the class of normalised, strongly regular matrices. Proof M = ( m ij ) , 0 < K � n ... an instance of BANDWIDTH. Let A = ( a ij ) 2 Z n � n : 8 � K if m ij = 1 < a ij = � n if m ij = 0 , i 6 = j : 0 if i = j A is normalised, strongly regular P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Set b = ( 1 , ..., n ) T P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Set b = ( 1 , ..., n ) T The answer to IPEV for A and b is "yes" ( ) 9 π 2 P n : a π ( i ) , π ( j ) � i � j for all i , j 2 N P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Set b = ( 1 , ..., n ) T The answer to IPEV for A and b is "yes" ( ) 9 π 2 P n : a π ( i ) , π ( j ) � i � j for all i , j 2 N ( ) � K � i � j if m π ( i ) π ( j ) = 1 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Set b = ( 1 , ..., n ) T The answer to IPEV for A and b is "yes" ( ) 9 π 2 P n : a π ( i ) , π ( j ) � i � j for all i , j 2 N ( ) � K � i � j if m π ( i ) π ( j ) = 1 ( ) K � j i � j j if m π ( i ) π ( j ) = 1 P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

The NP-completeness result Set b = ( 1 , ..., n ) T The answer to IPEV for A and b is "yes" ( ) 9 π 2 P n : a π ( i ) , π ( j ) � i � j for all i , j 2 N ( ) � K � i � j if m π ( i ) π ( j ) = 1 ( ) K � j i � j j if m π ( i ) π ( j ) = 1 ( ) "Yes" to BANDWIDTH P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity m are called v 1 , ..., v n 2 R WLD i¤ for some k and α j 2 R v k = ∑ � j 6 = k α j � v j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity m are called v 1 , ..., v n 2 R WLD i¤ for some k and α j 2 R v k = ∑ � j 6 = k α j � v j Gondran-Minoux LD i¤ for some U , V � f 1 , ..., n g , U \ V = ∅ , U , V 6 = ∅ and α j 2 R � j 2 U α j � v j = ∑ � ∑ j 2 V α j � v j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity m are called v 1 , ..., v n 2 R WLD i¤ for some k and α j 2 R v k = ∑ � j 6 = k α j � v j Gondran-Minoux LD i¤ for some U , V � f 1 , ..., n g , U \ V = ∅ , U , V 6 = ∅ and α j 2 R j 2 U α j � v j = ∑ � � ∑ j 2 V α j � v j Strongly LD i¤ � ∑ j = 1 ,..., n v j � x j = v does not have a unique solution for any v 2 R m P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity m are called v 1 , ..., v n 2 R WLD i¤ for some k and α j 2 R v k = ∑ � j 6 = k α j � v j Gondran-Minoux LD i¤ for some U , V � f 1 , ..., n g , U \ V = ∅ , U , V 6 = ∅ and α j 2 R � j 2 U α j � v j = ∑ � ∑ j 2 V α j � v j Strongly LD i¤ � ∑ j = 1 ,..., n v j � x j = v does not have a unique solution for any v 2 R m SLI = ) GMLI = ) WLI P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity A = ( a ij ) 2 R n � n , A = ( A 1 , ..., A n ) P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity A = ( a ij ) 2 R n � n , A = ( A 1 , ..., A n ) A is called: P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity A = ( a ij ) 2 R n � n , A = ( A 1 , ..., A n ) A is called: W eakly regular i¤ the following is not true for any k and α j 2 R : � A k = ∑ j 6 = k α j � A j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)

More on regularity A = ( a ij ) 2 R n � n , A = ( A 1 , ..., A n ) A is called: W eakly regular i¤ the following is not true for any k and α j 2 R : � A k = ∑ j 6 = k α j � A j Gondran-Minoux regular i¤ the following is not true for any U , V � f 1 , ..., n g , U \ V = ∅ , U , V 6 = ∅ and α j 2 R : � � j 2 U α j � A j = ∑ ∑ j 2 V α j � A j P.Butkoviµ c University of Birmingham Permuted eigenvector (Manchester 20 May 2008)