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On the curious commutativity of AMPD matrices Adhemar Bultheel Dept. Computer Science, KU Leuven Leipzig, 15 February 2018 http://nalag.cs.kuleuven.be/papers/ade/LEIPZIG18 Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15

On the curious commutativity of AMPD matrices Adhemar Bultheel Dept. Computer Science, KU Leuven Leipzig, 15 February 2018 http://nalag.cs.kuleuven.be/papers/ade/LEIPZIG18 Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 1 / 28
Definition AMPD = AM + D A , D ∈ C n × n diagonal matrices → � M = M π = G k = G π 1 G π 2 · · · G π n , k ∈ π π = ( π 1 , π 2 , . . . , π n ) a permutation of (1 , 2 , . . . , n )   I k − 1   α k β k   G k =  , k = 1 , . . . , n − 1  γ k δ k I n − k − 1 � I n − 1 � G n = . α n Note that G i and G j if | i − j | ≥ 2. But G k G k +1 � = G k +1 G k in general. Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 2 / 28
Definition AMPD = AM + D A , D ∈ C n × n diagonal matrices → � M = M π = G k = G π 1 G π 2 · · · G π n , k ∈ π π = ( π 1 , π 2 , . . . , π n ) a permutation of (1 , 2 , . . . , n )   I k − 1   α k β k   G k =  , k = 1 , . . . , n − 1  γ k δ k I n − k − 1 � I n − 1 � G n = . α n Note that G i and G j if | i − j | ≥ 2. But G k G k +1 � = G k +1 G k in general. Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 2 / 28
Definition AMPD = AM + D A , D ∈ C n × n diagonal matrices → � M = M π = G k = G π 1 G π 2 · · · G π n , k ∈ π π = ( π 1 , π 2 , . . . , π n ) a permutation of (1 , 2 , . . . , n )   I k − 1   α k β k   G k =  , k = 1 , . . . , n − 1  γ k δ k I n − k − 1 � I n − 1 � G n = . α n Note that G i and G j if | i − j | ≥ 2. But G k G k +1 � = G k +1 G k in general. Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 2 / 28
Definition AMPD = AM + D A , D ∈ C n × n diagonal matrices → � M = M π = G k = G π 1 G π 2 · · · G π n , k ∈ π π = ( π 1 , π 2 , . . . , π n ) a permutation of (1 , 2 , . . . , n )   I k − 1   α k β k   G k =  , k = 1 , . . . , n − 1  γ k δ k I n − k − 1 � I n − 1 � G n = . α n Note that G i and G j if | i − j | ≥ 2. But G k G k +1 � = G k +1 G k in general. Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 2 / 28
Definition AMPD = AM + D A , D ∈ C n × n diagonal matrices → � M = M π = G k = G π 1 G π 2 · · · G π n , k ∈ π π = ( π 1 , π 2 , . . . , π n ) a permutation of (1 , 2 , . . . , n )   I k − 1   α k β k   G k =  , k = 1 , . . . , n − 1  γ k δ k I n − k − 1 � I n − 1 � G n = . α n Note that G i and G j if | i − j | ≥ 2. But G k G k +1 � = G k +1 G k in general. Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 2 / 28
Table of contents Definition AMPD and observation The proof Unitary case The origin OPUC Generalization (ORFUC) Rational AMPD = RAMPD Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 3 / 28
Example Example 2     0 1 0 0 α 1 β 1   , G 2 =   , G 1 = γ 1 δ 1 0 0 α 2 β 2 0 0 1 0 γ 2 δ 2     α 1 β 1 α 2 β 1 β 2 α 1 β 1 0   � = G 2 G 1 =   G 1 G 2 = γ 1 δ 1 α 2 δ 1 β 2 α 2 γ 1 α 2 δ 1 β 2 0 γ 2 δ 2 γ 2 γ 1 γ 2 δ 1 δ 2 BUT det( G 1 G 2 ) = det( G 2 G 1 ) Thus also det( AG 1 G 2 + D ) = det( AG 2 G 1 + D ) D → D − λ I ⇒ σ ( AG 1 G 1 + D ) = σ ( AG 2 G 1 + D ) Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 4 / 28
Example Example 2     0 1 0 0 α 1 β 1   , G 2 =   , G 1 = γ 1 δ 1 0 0 α 2 β 2 0 0 1 0 γ 2 δ 2     α 1 β 1 α 2 β 1 β 2 α 1 β 1 0   � = G 2 G 1 =   G 1 G 2 = γ 1 δ 1 α 2 δ 1 β 2 α 2 γ 1 α 2 δ 1 β 2 0 γ 2 δ 2 γ 2 γ 1 γ 2 δ 1 δ 2 BUT det( G 1 G 2 ) = det( G 2 G 1 ) Thus also det( AG 1 G 2 + D ) = det( AG 2 G 1 + D ) D → D − λ I ⇒ σ ( AG 1 G 1 + D ) = σ ( AG 2 G 1 + D ) Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 4 / 28
Example Example 2     0 1 0 0 α 1 β 1   , G 2 =   , G 1 = γ 1 δ 1 0 0 α 2 β 2 0 0 1 0 γ 2 δ 2     α 1 β 1 α 2 β 1 β 2 α 1 β 1 0   � = G 2 G 1 =   G 1 G 2 = γ 1 δ 1 α 2 δ 1 β 2 α 2 γ 1 α 2 δ 1 β 2 0 γ 2 δ 2 γ 2 γ 1 γ 2 δ 1 δ 2 BUT det( G 1 G 2 ) = det( G 2 G 1 ) Thus also det( AG 1 G 2 + D ) = det( AG 2 G 1 + D ) D → D − λ I ⇒ σ ( AG 1 G 1 + D ) = σ ( AG 2 G 1 + D ) Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 4 / 28
Example Example 2     0 1 0 0 α 1 β 1   , G 2 =   , G 1 = γ 1 δ 1 0 0 α 2 β 2 0 0 1 0 γ 2 δ 2     α 1 β 1 α 2 β 1 β 2 α 1 β 1 0   � = G 2 G 1 =   G 1 G 2 = γ 1 δ 1 α 2 δ 1 β 2 α 2 γ 1 α 2 δ 1 β 2 0 γ 2 δ 2 γ 2 γ 1 γ 2 δ 1 δ 2 BUT det( G 1 G 2 ) = det( G 2 G 1 ) Thus also det( AG 1 G 2 + D ) = det( AG 2 G 1 + D ) D → D − λ I ⇒ σ ( AG 1 G 1 + D ) = σ ( AG 2 G 1 + D ) Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 4 / 28
Example Example 2     0 1 0 0 α 1 β 1   , G 2 =   , G 1 = γ 1 δ 1 0 0 α 2 β 2 0 0 1 0 γ 2 δ 2     α 1 β 1 α 2 β 1 β 2 α 1 β 1 0   � = G 2 G 1 =   G 1 G 2 = γ 1 δ 1 α 2 δ 1 β 2 α 2 γ 1 α 2 δ 1 β 2 0 γ 2 δ 2 γ 2 γ 1 γ 2 δ 1 δ 2 BUT det( G 1 G 2 ) = det( G 2 G 1 ) Thus also det( AG 1 G 2 + D ) = det( AG 2 G 1 + D ) D → D − λ I ⇒ σ ( AG 1 G 1 + D ) = σ ( AG 2 G 1 + D ) Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 4 / 28
Question Question Is in general σ ( AM π + D ) independent of π ? Let’s do some experiments ... and the result is.... Now prove it!!! Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 5 / 28
Question Question Is in general σ ( AM π + D ) independent of π ? Let’s do some experiments ... and the result is.... Now prove it!!! Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 5 / 28
Question Question Is in general σ ( AM π + D ) independent of π ? Let’s do some experiments ... and the result is.... Now prove it!!! Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 5 / 28
Question Question Is in general σ ( AM π + D ) independent of π ? Let’s do some experiments ... and the result is.... Now prove it!!! Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 5 / 28
Question Question Is in general σ ( AM π + D ) independent of π ? Let’s do some experiments ... and the result is.... Now prove it!!! Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 5 / 28
What does M π look like? The matrix M is the product of a number of G -matrices � � � � � � � � � � � � � � � � � � M 1 = or M 2 = or M 3 = � � � � � � � � � � � � � � � � � � M 1 = upper Hessenberg matrix π = (1 , 2 , 3 , 4 , 5 , 6) M 1 : π = (2 , 1 , 3 , 4 , 6 , 5) or (2 , 3 , 1 , 6 , 4 , 5) or ... M 3 = CMV matrix π = (1 , 3 , 5 , 2 , 4 , 6) or (5 , 1 , 3 , 4 , 2 , 6), or ... Adhemar Bultheel (KU leuven) Curious commutativity Leipzig, 15 February 2018 6 / 28

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