Number of Linear Transformations: Probability, Quantum Mechanics and Beyond

Algebra Level 5

The following discussion is just for the purpose of motivation and not strictly necessary to solve the problem. The actual problem is stated after the discussion.

Consider a system, which may be in any one of N N finitely many states S = { s i } i = 1 N \mathcal{S}=\{\vec{s}_i\}_{i=1}^{N} at a time. The system may be a classical one, e.g. weather of a city (here the states may be rainy or sunny ) or a quantum mechanical one, e.g. an atom (where the states may be its energy levels). The system may be acted upon by a transformation T : S S \mathcal{T} : \mathcal{S} \to \mathcal{S} (e.g. performing a precipitation measurement, or exposing the atom to a radiation field), which changes its state according to some rule.

We are seeking to develop a linear theory for this general system. This means that a general state ζ \zeta of the system may be represented as a real linear combination of the states S \mathcal{S} ζ = i β i s i \zeta = \sum_i \beta_i \vec{s}_i for some real numbers β i \beta_i , and the transformation T \mathcal{T} may be represented by some real N × N N \times N matrix T T , which acts upon the vector ζ \zeta , according to the rules of linear algebra.

We now need to relate the probability of finding the system in a particular state s i \vec{s}_i with its state vector β \vec{\beta} . Note that the total probability of different states before and after the measurement must be equal to 1. We are aware of two ways of doing that:

(1) Classical Probability Theory : Constrain the vector β \vec{\beta} to be elementwise non-negative with i β i = 1. \sum_i \beta_i=1. Then we are in the realm of classical probability theory (Markov chain) with the probability of s i \vec{s}_i being simply the corresponding β i \beta_i and the transformation matrix T T can be any N × N N\times N stochastic matrix , since it preserves the 1 1 -norm.

(2) Quantum Mechanics : Let the vector β \vec{\beta} to be any N N dimensional real vector with the constraint i β i 2 = 1. \sum_i \beta_i^2 =1. Then we are in the realm of quantum mechanics, with the probability of s i \vec{s}_i being β i 2 \beta_i^2 . The transformation matrix T T could be any N × N N\times N orthogonal matrix, since it preserves the 2 2 -norm.

A natural question arises: Can we continue this procedure (and derive new, meaningful theory) with any other norm? Of course, for a theory to be meaningful, we need to have access to lots of possible measurements, i.e. legitimate T T matrices (we have infinitely many such feasible T T matrices for the above two cases).

To be specific, consider the 4 4 -norm . Then, the probability of finding the system ζ \zeta in state s i s_i would be β i 4 \beta_i^4 .


The actual problem

Consider the set of all N × N N \times N real invertible matrices T N \mathcal{T}_N such that, upon multiplication, it preserves the 4 th 4^\text{th} norm of any real N N dimensional vector. In other words, the matrix T T N T \in \mathcal{T}_N iff T x 4 = 1 ||T\vec{x}||_4=1 for all x R N \vec{x} \in \mathbb{R}^N with x 4 = 1 ||x||_4=1 . This means that if we denote the N N dimensional vector y = T x \vec{y}= T \vec{x} , then we require i y i 4 = 1 \sum_i y_i^4 = 1 for all x R N \vec{x} \in \mathbb{R}^N such that i x i 4 = 1 \sum_i x_i^4=1 .

Problem : What is the cardinality of T N \mathcal{T}_N for N = 10 ? N=10?

If you think the answer is infinite (like above), enter 1 -1 as your answer.

Picture Courtesy: Nataly Meerson


The answer is 3715891200.

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1 solution

Abhishek Sinha
Dec 27, 2016

Let T = [ t i j ] N × N T= [t_{ij}]_{N\times N} . Since it preserves the fourth norm of the unit vectors, we must have i t i j 4 = 1 , j . ( 1 ) \sum_{i}t_{ij}^4 =1, \hspace{10pt} \forall j. \hspace{10 pt} (1) Now we will show that for any i i and j k j \neq k , we have t i j 2 t i k 2 = 0 t_{ij}^2 t_{ik}^2=0 . For this, consider the vector x \vec{x} whose all elements excepting j j th and k k th elements are zero and x j = x k = 1 2 1 / 4 x_j=x_k= \frac{1}{2^{1/4}} , so that x 4 = 1 ||\vec{x}||_4=1 . Now consider the i i th element of T x T\vec{x} . It is given by ( T x ) i = 1 2 1 / 4 ( t i j + t i k ) (T\vec{x})_i= \frac{1}{2^{1/4}}(t_{ij}+ t_{ik}) . By the norm-preserving property of T T , we require 1 2 i ( t i j + t i k ) 4 = 1 \frac{1}{2}\sum_i (t_{ij}+t_{ik})^4=1 Expanding binomially, the i i th term of the above sum becomes t i j 4 + 4 t i j 3 t i k + 6 t i j 2 t i k 2 + 4 t i j t i k 3 + t i k 4 = 2 t_{ij}^4+ 4t_{ij}^3t_{ik} + 6 t_{ij}^2t_{ik}^2 + 4 t_{ij}t_{ik}^3+ t_{ik}^4=2 Summing over i i and using (1), we obtain i 4 t i j 3 t i k + 6 t i j 2 t i k 2 + 4 t i j t i k 3 = 0 , ( a ) \sum_i 4t_{ij}^3t_{ik} + 6 t_{ij}^2t_{ik}^2 + 4 t_{ij}t_{ik}^3 = 0, \hspace{10pt} (a) Similarly, taking x j = x k = 1 2 1 / 4 x_j= -x_k= \frac{1}{2^{1/4}} , we obtain i 4 t i j 3 t i k + 6 t i j 2 t i k 2 4 t i j t i k 3 = 0 , ( b ) \sum_i -4t_{ij}^3t_{ik} + 6 t_{ij}^2t_{ik}^2 - 4 t_{ij}t_{ik}^3 = 0, \hspace{10pt} (b) Summing (a) and (b), we obtain i t i j 2 t i k 2 = 0 \sum_i t_{ij}^2t_{ik}^2 =0 Since the matrix T T is real, we conclude t i j 2 t i k 2 = 0 , i t_{ij}^2t_{ik}^2=0, \forall i . This implies that there is at most one non-zero element at each row of T T . On the other hand, due to the requirement of invertibility of T T , we need at least one non-zero element per row. This, along with the condition (1) implies that the set of all feasible T N \mathcal{T}_N is the generalized permutation matrices , where the non-zero element 1 1 could have either positive or negative sign. Hence, there are exactly 2 N N ! 2^{N} N! of such linear transformations.

Discussion It can be easily seen that the proof easily extends beyond p = 4 p=4 . Since in this case, the only feasible transformations are rather trivial (essentially shuffling and switching the direction of bases), we do not obtain any useful physical theory in this way, for any norm other than p = 1 p=1 (Probability theory) and p = 2 p=2 (Quantum Mechanics), the two main pillars of modern mathematics and physics.

Isn't the fact that the transform is invertible follow from the fact the transform preserves the 4 norm?

Brilliant question by the way.

A Former Brilliant Member - 4 years, 5 months ago

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Thanks for pointing it out. Of course, the invertibility requirement is superfluous. It can be shown that any norm-preserving transformation is invertible. To show this, let T T be a norm-preserving transform such that T x = T y Tx=Ty i.e, T ( x y ) = 0 T(x-y)=\vec{0} Hence, T ( x y ) p = 0 ||T(x-y)||_p = 0 By the norm-preserving property of T T , the above implies x y p = 0 ||x-y||_p = 0 This implies that x = y x=y , hence the matrix T T is invertible.

Abhishek Sinha - 4 years, 5 months ago

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Exactly! In fact the same proof works for any norm (not just a p norm).

A Former Brilliant Member - 4 years, 5 months ago

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