The number of eigenvalues of a tensor Dustin Cartwright 1 Yale - PowerPoint PPT Presentation

The number of eigenvalues of a tensor Dustin Cartwright 1 Yale University/MPIM Bonn May 31, 2012 1 joint with Bernd Sturmfels Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 1 / 11

Eigenvalues of a tensor A = ( a i 1 ,..., i m ): order- m tensor of size n × · · · × n with entries in C Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 2 / 11

Eigenvalues of a tensor A = ( a i 1 ,..., i m ): order- m tensor of size n × · · · × n with entries in C From this we have the operator on x ∈ C n written Ax m − 1 and defined by n n � � ( Ax m − 1 ) j = · · · a ji 2 ··· i m x i 2 · · · x i m i 2 =1 i m =1 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 2 / 11

Eigenvalues of a tensor A = ( a i 1 ,..., i m ): order- m tensor of size n × · · · × n with entries in C From this we have the operator on x ∈ C n written Ax m − 1 and defined by n n � � ( Ax m − 1 ) j = · · · a ji 2 ··· i m x i 2 · · · x i m i 2 =1 i m =1 Definition (Qi, Lim) The (E-)eigenvectors of A are the fixed points (up to scaling) of this operator: Ax m − 1 = λ x for x � = 0 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 2 / 11

Eigenvalues of a tensor A = ( a i 1 ,..., i m ): order- m tensor of size n × · · · × n with entries in C From this we have the operator on x ∈ C n written Ax m − 1 and defined by n n � � ( Ax m − 1 ) j = · · · a ji 2 ··· i m x i 2 · · · x i m i 2 =1 i m =1 Definition (Qi, Lim) The (E-)eigenvectors of A are the fixed points (up to scaling) of this operator: Ax m − 1 = λ x for x � = 0 We will call ( λ, x ) an eigenpair and consider them up to the equivalence: ( λ, x ) ∼ ( t m − 2 λ, tx ) for any t � = 0 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 2 / 11

The number of eigenpairs Theorem (Ni-Qi-Wang-Wang, C-Sturmfels) The number of equivalence classes of eigenpairs of A is either infinity or ( m − 1) n − 1 n − 1 � ( m − 1) i = (1) m − 2 i =0 when counted with multiplicity. For a generic tensor, there are exactly this many equivalence classes of eigenpairs. Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 3 / 11

The number of eigenpairs Theorem (Ni-Qi-Wang-Wang, C-Sturmfels) The number of equivalence classes of eigenpairs of A is either infinity or ( m − 1) n − 1 n − 1 � ( m − 1) i = (1) m − 2 i =0 when counted with multiplicity. For a generic tensor, there are exactly this many equivalence classes of eigenpairs. Given the first sentence, the second can be seen by considering the diagonal tensor � 1 if i 1 = · · · = i m A i 1 ,..., i m = 0 otherwise . It has exactly as many eigenvalues up to equivalence as the quantity in (1). Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 3 / 11

A proof for generic tensors Ax m − 1 = λ x Substitute λ = µ m − 2 : Ax m − 1 = µ m − 2 x (2) n equations, homogeneous of degree m − 1 in µ, x 1 , . . . , x n . Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A proof for generic tensors Ax m − 1 = λ x Substitute λ = µ m − 2 : Ax m − 1 = µ m − 2 x (2) n equations, homogeneous of degree m − 1 in µ, x 1 , . . . , x n . ezout’s theorem, these have ( m − 1) n solutions up to scaling. By B´ Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A proof for generic tensors Ax m − 1 = λ x Substitute λ = µ m − 2 : Ax m − 1 = µ m − 2 x (2) n equations, homogeneous of degree m − 1 in µ, x 1 , . . . , x n . ezout’s theorem, these have ( m − 1) n solutions up to scaling. By B´ One solution is trivial: ( µ : x 1 : . . . : x n ) = (1 : 0 : . . . : 0) An equivalence class of eigenpairs with λ � = 0 corresponds to m − 2 solutions of (2) by taking µ to be the m − 2 roots of λ . Thus, we get ( m − 1) n − 1 eigenpairs m − 2 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A proof for generic tensors Ax m − 1 = λ x Substitute λ = µ m − 2 : Ax m − 1 = µ m − 2 x (2) n equations, homogeneous of degree m − 1 in µ, x 1 , . . . , x n . ezout’s theorem, these have ( m − 1) n solutions up to scaling. By B´ One solution is trivial: ( µ : x 1 : . . . : x n ) = (1 : 0 : . . . : 0) An equivalence class of eigenpairs with λ � = 0 corresponds to m − 2 solutions of (2) by taking µ to be the m − 2 roots of λ . Thus, we get ( m − 1) n − 1 eigenpairs m − 2 Unfortunately, this approach cannot directly treat the case when λ = 0. Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A geometric reinterpretation � � P n = ( µ : x 1 : . . . : x n ) � = 0 / ( µ : x 1 : . . . : x n ) ∼ ( t µ : tx 1 : . . . tx n ) ↓ � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 5 / 11

A geometric reinterpretation Quotient of projective space by group which multiplies µ by e 2 π i / ( m − 2) . � � P n = ( µ : x 1 : . . . : x n ) � = 0 / ( µ : x 1 : . . . : x n ) ∼ ( t µ : tx 1 : . . . tx n ) ↓ � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 5 / 11

A geometric reinterpretation Quotient of projective space by group which multiplies µ by e 2 π i / ( m − 2) . � � P n = ( µ : x 1 : . . . : x n ) � = 0 / ( µ : x 1 : . . . : x n ) ∼ ( t µ : tx 1 : . . . tx n ) ↓ � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 P n to P n , used Summary of previous slide: lift the eigenvalue problem from � B´ ezout’s theorem, and the fact that the quotient is ( m − 2)-to-1 except: when x 1 = · · · = x n = 0 (subtract this out anyways) when λ = µ = 0 (as noted, exceptional from this perspective) Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 5 / 11

A geometric reinterpretation Quotient of projective space by group which multiplies µ by e 2 π i / ( m − 2) . � � P n = ( µ : x 1 : . . . : x n ) � = 0 / ( µ : x 1 : . . . : x n ) ∼ ( t µ : tx 1 : . . . tx n ) ↓ � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 P n to P n , used Summary of previous slide: lift the eigenvalue problem from � B´ ezout’s theorem, and the fact that the quotient is ( m − 2)-to-1 except: when x 1 = · · · = x n = 0 (subtract this out anyways) when λ = µ = 0 (as noted, exceptional from this perspective) Any quotient of a smooth variety is smooth where the group action is free. In this case, it so happens that the quotient is also smooth where µ = 0. Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 5 / 11

Intersections in � P n � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 The intersection of the eigenvalue equations in the rational Chow ring is ( m − 1) n m − 2 . (3) (not an integer if m > 3.) Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 6 / 11

Intersections in � P n � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 The intersection of the eigenvalue equations in the rational Chow ring is ( m − 1) n m − 2 . (3) (not an integer if m > 3.) This quantity can be computed as the normalized volume of a polytope, but also as the quotient B´ ezout number by the order of the group of ( m − 2)th roots of unity. Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 6 / 11

Intersections in � P n � � P n = � / ( λ : x 1 : . . . : x n ) ∼ ( t m − 2 λ : tx 1 : . . . : tx n ) ( λ : x 1 : . . . : x n ) � = 0 The intersection of the eigenvalue equations in the rational Chow ring is ( m − 1) n m − 2 . (3) (not an integer if m > 3.) This quantity can be computed as the normalized volume of a polytope, but also as the quotient B´ ezout number by the order of the group of ( m − 2)th roots of unity. The unique singular point x = 0 counts for 1 m − 2 so the number of equivalence classes of eigenvalues is ( m − 1) n 1 − m − 2 m − 2 Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 6 / 11

Corollaries Theorem (Ni-Qi-Wang-Wang, C-Sturmfels) The number of equivalence classes of eigenpairs of A is either infinity or n − 1 ( m − 1) n − 1 � ( m − 1) i = (1) m − 2 i =0 when counted with multiplicity. For a generic tensor, there are exactly this many equivalence classes of eigenpairs. Corollary For m ≥ 3 , the number of equivalence classes of eigenvalues grows exponentially in n. Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 7 / 11