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

The number of eigenvalues of a tensor

Dustin Cartwright1

Yale University/MPIM Bonn

May 31, 2012

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

Eigenvalues of a tensor

A = (ai1,...,im): 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 = (ai1,...,im): order-m tensor of size n × · · · × n with entries in C From this we have the operator on x ∈ Cn written Axm−1 and defined by (Axm−1)j =

n

• i2=1

· · ·

n

• im=1

aji2···imxi2 · · · xim

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 2 / 11

Eigenvalues of a tensor

A = (ai1,...,im): order-m tensor of size n × · · · × n with entries in C From this we have the operator on x ∈ Cn written Axm−1 and defined by (Axm−1)j =

n

• i2=1

· · ·

n

• im=1

aji2···imxi2 · · · xim Definition (Qi, Lim) The (E-)eigenvectors of A are the fixed points (up to scaling) of this

• perator:

Axm−1 = λx for x = 0

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 2 / 11

Eigenvalues of a tensor

A = (ai1,...,im): order-m tensor of size n × · · · × n with entries in C From this we have the operator on x ∈ Cn written Axm−1 and defined by (Axm−1)j =

n

• i2=1

· · ·

n

• im=1

aji2···imxi2 · · · xim Definition (Qi, Lim) The (E-)eigenvectors of A are the fixed points (up to scaling) of this

• perator:

Axm−1 = λx for x = 0 We will call (λ, x) an eigenpair and consider them up to the equivalence: (λ, x) ∼ (tm−2λ, 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 m − 2 =

n−1

• i=0

(m − 1)i (1) 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 m − 2 =

n−1

• i=0

(m − 1)i (1) 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 Ai1,...,im =

• 1

if i1 = · · · = im

• therwise.

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

Axm−1 = λx Substitute λ = µm−2: Axm−1 = µm−2x (2) n equations, homogeneous of degree m − 1 in µ, x1, . . . , xn.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A proof for generic tensors

Axm−1 = λx Substitute λ = µm−2: Axm−1 = µm−2x (2) n equations, homogeneous of degree m − 1 in µ, x1, . . . , xn. By B´ ezout’s theorem, these have (m − 1)n solutions up to scaling.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A proof for generic tensors

Axm−1 = λx Substitute λ = µm−2: Axm−1 = µm−2x (2) n equations, homogeneous of degree m − 1 in µ, x1, . . . , xn. By B´ ezout’s theorem, these have (m − 1)n solutions up to scaling. One solution is trivial: (µ : x1 : . . . : xn) = (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 m − 2 eigenpairs

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 4 / 11

A proof for generic tensors

Axm−1 = λx Substitute λ = µm−2: Axm−1 = µm−2x (2) n equations, homogeneous of degree m − 1 in µ, x1, . . . , xn. By B´ ezout’s theorem, these have (m − 1)n solutions up to scaling. One solution is trivial: (µ : x1 : . . . : xn) = (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 m − 2 eigenpairs 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

Pn =

• (µ : x1 : . . . : xn) = 0
• /(µ : x1 : . . . : xn) ∼ (tµ : tx1 : . . . txn)

↓

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

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 e2πi/(m−2). Pn =

• (µ : x1 : . . . : xn) = 0
• /(µ : x1 : . . . : xn) ∼ (tµ : tx1 : . . . txn)

↓

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

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 e2πi/(m−2). Pn =

• (µ : x1 : . . . : xn) = 0
• /(µ : x1 : . . . : xn) ∼ (tµ : tx1 : . . . txn)

↓

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

Summary of previous slide: lift the eigenvalue problem from Pn to Pn, used B´ ezout’s theorem, and the fact that the quotient is (m − 2)-to-1 except: when x1 = · · · = xn = 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 e2πi/(m−2). Pn =

• (µ : x1 : . . . : xn) = 0
• /(µ : x1 : . . . : xn) ∼ (tµ : tx1 : . . . txn)

↓

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

Summary of previous slide: lift the eigenvalue problem from Pn to Pn, used B´ ezout’s theorem, and the fact that the quotient is (m − 2)-to-1 except: when x1 = · · · = xn = 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 Pn

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

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 Pn

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

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 Pn

• Pn =
• (λ : x1 : . . . : xn) = 0
• /(λ : x1 : . . . : xn) ∼ (tm−2λ : tx1 : . . . : txn)

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 m − 2 − 1 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 (m − 1)n − 1 m − 2 =

n−1

• i=0

(m − 1)i (1) 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

Corollaries

Theorem (Ni-Qi-Wang-Wang, C-Sturmfels) The number of equivalence classes of eigenpairs of A is either infinity or (m − 1)n − 1 m − 2 =

n−1

• i=0

(m − 1)i (1) when counted with multiplicity. For a generic tensor, there are exactly this many equivalence classes of eigenpairs. Corollary If the entries of A are real and either m or n is odd, then A has at least

• ne real eigenpair.

If m or n is odd, then the sum in (1) is odd.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 7 / 11

Corollaries

Theorem (Ni-Qi-Wang-Wang, C-Sturmfels) The number of equivalence classes of eigenpairs of A is either infinity or (m − 1)n − 1 m − 2 =

n−1

• i=0

(m − 1)i (1) when counted with multiplicity. For a generic tensor, there are exactly this many equivalence classes of eigenpairs. Corollary The characteristic polynomial φA(λ) has degree (m − 1)n − 1 m − 2 =

n−1

• i=0

(m − 1)i

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 7 / 11

The characteristic polynomial

The coefficients of the characteristic polynomial φA(λ) are polynomials in the entries of the tensor. It vanishes on eigenvalues whose eigenvector has been normalized to have x · x = 1.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 8 / 11

The characteristic polynomial

The coefficients of the characteristic polynomial φA(λ) are polynomials in the entries of the tensor. It vanishes on eigenvalues whose eigenvector has been normalized to have x · x = 1. More precisely: let I ⊂ C[ai1,...,im, xj, λ] be the ideal generated by: λx1 −

n

• i2=1

· · ·

n

• im=1

a1i2···imxi2 · · · xin . . . λxn −

n

• i2=1

· · ·

n

• im=1

ani2···imxi2 · · · xin x2

1 + · · · + x2 n − 1

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 8 / 11

The characteristic polynomial

The coefficients of the characteristic polynomial φA(λ) are polynomials in the entries of the tensor. It vanishes on eigenvalues whose eigenvector has been normalized to have x · x = 1. More precisely: let I ⊂ C[ai1,...,im, xj, λ] be the ideal generated by: λx1 −

n

• i2=1

· · ·

n

• im=1

a1i2···imxi2 · · · xin . . . λxn −

n

• i2=1

· · ·

n

• im=1

ani2···imxi2 · · · xin x2

1 + · · · + x2 n − 1

The elimination ideal I ∩ C[ai1,...,im, λ] is a principal ideal. If m is even, characteristic polynomial φA(λ) is the the generator of this ideal. If m is

• dd, the generator is φA(λ2).

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 8 / 11

The roots of the characteristic polynomial

Normalized eigenvalues are roots of the characteristic polynomial but not necessarily conversely. Example Let A be the 2 × 2 × 2 tensor with a111 = a221 = 1 and a122 = a222 = √ −1 and 0 entries elsewhere. The eigenvalue problem for A is: x2

1 + ix1x2 = λx1

and x1x2 + ix2

2 = λx2.

This has a normalized eigenvalue of λ if and only λ = 0. Therefore, the characteristic polynomial is identically zero. Thus φA(0) = 0, even though λ = 0 is not a (normalized) eigenvalue.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 9 / 11

The roots of the characteristic polynomial

Example Let A be the symmetric 2 × 2 × 2 tensor with a111 = −2i a112 = a121 = a211 = 1 a222 = 1 a122 = a212 = a221 = 0 Then the characteristic polynomial of A vanishes identically, but A has

• nly a single normalized eigenvalue, namely 1.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 10 / 11

The roots of the characteristic polynomial

Example Let A be the symmetric 2 × 2 × 2 tensor with a111 = −2i a112 = a121 = a211 = 1 a222 = 1 a122 = a212 = a221 = 0 Then the characteristic polynomial of A vanishes identically, but A has

• nly a single normalized eigenvalue, namely 1.

Why? Because A also has a non-normalized eigenpair with eigenvalue 1 and eigenvector (1, i). For small perturbations of A, this can take on any value as a normalized eigenvalue.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 10 / 11

Symmetric tensors

A tensor is symmetric if it is invariant under all permutations of the m factors. Theorem If A is symmetric, then A has at most (m − 1)n − 1 m − 2 distinct normalized eigenvalues.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 11 / 11

Symmetric tensors

A tensor is symmetric if it is invariant under all permutations of the m factors. Theorem If A is symmetric, then A has at most (m − 1)n − 1 m − 2 distinct normalized eigenvalues. Two caveats: As in the example on the previous slide, the characteristic polynomial

• f A may still vanish identically.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 11 / 11

Symmetric tensors

A tensor is symmetric if it is invariant under all permutations of the m factors. Theorem If A is symmetric, then A has at most (m − 1)n − 1 m − 2 distinct normalized eigenvalues. Two caveats: As in the example on the previous slide, the characteristic polynomial

• f A may still vanish identically.

There may also be infinitely many eigenvectors with the same normalized eigenvalue, and with a different method of normalization, these may yield infinitely many eigenvalues.

Dustin Cartwright (Yale/MPIM) The number of eigenvalues of a tensor May 31, 2012 11 / 11