 # A Unified Theory for Tensor Ranks and its Application

In this paper, we present a unified theory for tensor ranks such that they are natural extension of matrix ranks. We present some axioms for tensor rank functions. The CP rank, the max-Tucker rank and the submax-Tucker rank are tensor rank functions. The CP rank is subadditive but not proper. The max-Tucker rank naturally arises from the Tucker decomposition. It is proper and subadditive, but not strongly proper. The submax-Tucker rank is also associated with the Tucker decomposition, but is a new tensor rank function. It is strongly proper but not subadditive. We define a partial order among tensor rank functions and show that there exists a unique smallest tensor rank function. The CP rank, and the max-Tucker rank are not the smallest tensor rank function. We define the closure of a strongly proper tensor rank function, and show that it is also a strongly proper tensor rank function. A strongly proper tensor rank function is closed if it is equal to its closure. We show that the smallest tensor rank function is strongly proper and closed. Our theoretic analysis indicates that the submax-Tucker rank is a good choice for low rank tensor approximation and tensor completion. An application of the submax-Tucker rank is presented.

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## 1 Introduction

Tensor ranks play a crucial role in low rank tensor approximation, tensor completion and tensor recovery [1, 6, 7, 8, 9, 11, 13, 14, 15, 16, 17, 18]. However, its theory is still not matured yet. Two most popular tensor ranks are the CP rank and the Tucker rank . Are there an axiom system to define tensor ranks such that they are natural extensions of matrix ranks? Are there some relations among these tensor ranks? Are there some smallest tensor ranks satisfying such axiom systems? Is there a unified theory for tensor ranks? These are the research topics of this paper.

The set of all nonnegative integers is denoted by . The set of all positive integers is denoted by . Let . Denote the set of all real th order tensors of dimension by . If , then we denote it by . Here “CT” means cubic tensors. Denote the set of all real tensors by

. Thus, scalars, vectors, matrices are a part of

. Let . We call a rank-one tensor if and only there are nonzero vectors for , such that

 X=x(1)∘⋯∘x(m).

Here, is the tensor outer product. Then, nonzero vectors and scalars are all rank-one tensors in this sense.

Suppose that and . Let and for . Suppose that we have with if for . Then we say that is a subtensor of . If furthermore, all the entries of , which are not in , are zero, then we say that is an essential subtensor of .

In the next section, we present a set of axioms for tensor rank functions. We list six properties which are essential for tensor ranks. We also list some properties satisfied by some tensor rank functions. In particular, we define a partial order “” among tensor rank functions, and show that there exists a unique smallest tensor rank function . It is interesting to identify this smallest tensor rank function and its properties.

We study the CP rank and the Tucker rank in Section 3. The Tucker rank is a vector rank. We derive two tensor rank functions from this, and call them the max-Tucker rank and the submax-Tucker rank respectively. We show that the CP rank, the max-Tucker rank and the submax-Tucker rank are all tensor rank functions. The max-Tucker rank naturally arises from the Tucker decomposition. The submax-Tucker rank is also associated with the Tucker decomposition, but is a new tensor rank function introduced in this paper. The CP rank is subadditive but not proper. The max-Tucker rank is proper, subadditive but not strongly proper. The submax-Tucker rank is strongly proper but not subadditive.

We define the closure of a strongly proper tensor rank function in Section 4, and show that it is also a strongly proper tensor rank function. A strongly proper tensor rank function is closed if and only if it is equal to its closure. We also show that is strongly proper and closed.

Our theoretic analysis indicates that the submax-Tucker rank is a good choice for low rank tensor approximation and tensor completion. An application of the submax-Tucker rank in internet traffic data approximation is presented in Section 5.

Some final remarks are made in Section 6.

We use small letters to denote scalars, small bold letters to denote vectors, capital letters to denote matrices, and calligraphic letters to denote tensors.

## 2 Axioms and Properties of Tensor Rank Functions

Let . Consider . Suppose . An entry is called a diagonal entry of if . Otherwise, is called a diagonal entry of . If all the off-diagonal entries of are zero, then is called a diagonal tensor. If is diagonal, and all the diagonal entries of is , then is called the identity tensor of , and denoted as . Clearly, the identity tensor is unique to .

###### Definition 2.1

Suppose that . If satisfies the following six properties, then is called a tensor rank function.

Property 1 Suppose that . Then if and only is a zero tensor, and if and only if is a rank-one tensor.

Property 2 For all , .

Property 3 Let , . Then is equal to the matrix rank of the matrix corresponding to .

Property 4 Let , , and is a real nonzero number. Then .

Property 5 Let , , and is a permutation on . Then , where , .

Property 6 Let . Suppose that , and is a subtensor of . Then . If is an essential subtensor of , then .

These six properties are essential for tensor ranks. Property 1 specifies rank zero tensors and rank one tensors. Though the tensor rank theory is not matured, there are no arguments in rank zero and rank one tensors in the literature. Property 2 fixes the value of the tensor rank for identity tensors. Property 3 justifies the tensor rank is an extension of the matrix rank. Property 4 claims that the tensor rank is not changed when a tensor is multiplied by a nonzero real number. Property 5 says that the roles of the modes are balanced. Property 6 justifies the subtensor rank relation.

Suppose that are two tensor rank functions. If for any we always have , then we say that the tensor rank function is not greater than the tensor rank function and denote this relation as .

###### Theorem 2.2

Suppose that are two tensor rank functions. Define by

 r(X)=min{r1(X),r2(X)},

for any . Then is a tensor rank function, and .

Proof For any , let . Then Properties 1, 2, 3 and 4 hold clearly from the definition of tensor rank functions.

To show Property 5, we assume that is a permutated tensor of . Then and . Hence, and Property 5 is obtained.

Now we assume that is a subtensor of . Then and . Hence since and . For an essential subtensor , and . Hence and Property 6 holds.

Thus, we conclude that is a tensor rank function.

clearly, and .

.

###### Theorem 2.3

There exists a unique tensor rank function , such that for any tensor rank function , we have .

Proof For any , define . This is well-defined as tensor rank functions take values on . Now we show that is a tensor rank function.

1) Suppose is a zero tensor in . Then for any tensor rank function , . This implies that by the definition of . On the other hand, suppose that for some . Then for some tensor rank function , . Hence, is a zero tensor from Property 1 of the tensor rank function . Similarly, we may show that if and only if is a rank-one tensor.

2) For any , for all tensor rank functions . Thus .

3) Let . Let be the corresponding matrix in . Then for any tensor rank function , reduces to the matrix rank of . and hence all of are equal. Hence, will be the matrix rank of and Property 3 holds.

4) For any and any tensor rank function , for any . Thus, .

5) We have Properties 5 and 6 in a similar way as in the proof of Theorem 2.2 and omit the details here.

By the definition, for any tensor rank function .

Suppose that and are two tensor rank functions with the property that and for any tensor rank function. Then . We see that . Thus, such a tensor rank function is unique.

.

We call the smallest tensor rank function. How can we identify such a tensor rank function ? To identify the smallest tensor rank function and its properties is significant in theory and has potential values in applications.

The six properties in Definition 3.1 are essential to tensor rank functions. There are some other properties which are satisfied by some tensor rank functions.

###### Definition 2.4

Suppose that is a tensor rank function. We say that is a proper tensor rank function if for any and , we have .

###### Definition 2.5

Suppose that is a tensor rank function. We say that is a subadditive tensor rank function if for any , and , we have

 r(X+Y)≤r(X)+r(Y).
###### Proposition 2.6

Suppose that is a proper tensor rank function. Let with and . Then we have

 r(X)≤max{n1,⋯,nm}. (2.1)

Proof Let and with a subtensor . Then from Property 6. Together with since is proper, the result is arrived.

.

For with , we define submax as the second largest value of .

###### Definition 2.7

Suppose that is a tensor rank function. We say that is a strongly proper tensor rank function if for any with , and , we have

 r(X)≤submax{n1,⋯,nm}. (2.2)

We cannot change submax in (2.2) to the third largest value of as this violates Property 1 of Definition 2.1.

## 3 CP Rank, Max-Tucker Rank and Submax-Tucker Rank

We now study the CP rank .

###### Definition 3.1

Suppose that and . Suppose that there are for and such that

 X=r∑p=1a(i,p)∘⋯∘a(m,p), (3.3)

then we say that has a CP decomposition (3.3). The smallest integer such that (3.3) holds is called the CP rank of , and denoted as CPRank.

###### Theorem 3.2

The CP rank is a subadditive tensor rank function. It is not a proper tensor rank function.

Proof In this proof, is the CP rank. We first show that the CP rank is a tensor rank function. Properties 1, 3 and 4 hold clearly from the definition of the CP rank. Before we show Property 2, we can assert that for all since , where with the unique nonzero entry . In the following, we show Property 2 by induction for . We fix here.

For , reduces to the identity matrix and hence Property 2 is true for such a case. Now we assume that . Then we show .

Assume that with . Then

 Im,n=Im+1,n⋅en≡r∑p=1((en)Ta(m+1,p))a(1,p)∘⋯∘a(m,p).

This indicates that since . This contradicts the assumption that and hence .

Hence, Property 2 holds.

For Property 5, we have that if when is a permutation of with . Hence we have Property 5.

For property 6, assume that is a subtensor of . For , let and be a subtensor of by a similar way of from . Then we have that and since is rank-1 tensor. This means that .

Furthermore, let be an essential subtensor of . Clearly, since is a subtensor of . Now we assume that . Suppose that and let

 ¯a(i,p)={ali,i,if  (i,p)=(li,i)∈Tl0,Otherwise

where is the index set related to subtensor . Then and hence . Hence and Property 6 is satisfied and the CP rank is a tensor rank function.

Hence, the CP rank is a tensor rank function.

Suppose that with and . Let

 X=r1∑p=1a(1,p)∘⋯∘a(m,p),Y=r2∑q=1b(1,q)∘⋯∘b(m,q).

It holds that

 X+Y=r1∑p=1a(1,p)∘⋯∘a(m,p)+r2∑q=1b(1,q)∘⋯∘b(m,q).

Hence, . This shows that it is subadditive. By , the CP rank of a tensor given by Kruskal is between and . Thus, the CP rank is not a proper tensor rank function. .

We now study the Tucker rank. In some papers such as , the -rank is called the Tucker rank.

Suppose that and . We may unfold to a matrix for . Denote the matrix ranks of as for . Then the vector is called the -rank of .

The -rank is a vector rank. Hence it does not satisfy Definition 3.1. However, if we define

 r=max{r1,⋯,rm}, (3.4)

then we have the following proposition.

###### Theorem 3.3

The function defined by (3.4) is a proper, subadditive tensor rank function. But it is not strongly proper.

Proof We first show that rank function defined by (3.4) is a tensor rank function. To see this, it suffices to show that Property 1-6 are all satisfied.

1) Suppose that for is a zero tensor. Then are zero matrices for . This implies that for . By (3.4), we have . On the other hand, assume that for some with . This means that for , which means that and hence is a zero tensor.

Suppose that , then for all . This can be seen as follows. Assume that there exists such that , then since . From above analysis, if and only if . This contradicts with .

Let . Then From , we have that is rank one. From and , we have that for all is also rank one.

Similarly, we have that for any , is rank one.

Thus, for some and hence is a rank one tensor.

Conversely, if is a rank one tensor, then for some nonzero vectors . Then and for all . Thus .

Based on the above analysis, Property 1 is satisfied.

2) Denote Then is an identity matrix for , and hence . Thus, .

3) When , then , and for any . Clearly, and hence .

4) Suppose that . For any , and any , and hence . Hence .

5) Suppose that and is any permuted tensor of . Then will be for . So . Hence and the result holds.

6) Suppose that is a subtensor of . Then for all , will be a submatrix of and since is matrix rank. So . Suppose that is an essential subtensor of , then is an essential subtensor of and hence . So we can assert that .

Now we conclude that defined by (3.4) is a tensor rank function.

It is clear that such a tensor rank function is proper from its definition. Furthermore, we have that such rank is also subadditive since matrix rank is subadditive.

In addition, we consider with where is the identity matrix of three dimension. Hence . Hence we conclude that such a tensor rank function is not strongly proper.

.

Thus, we call this tensor rank function the max-Tucker rank in this paper, and denote it as max-TucRank.

Note that the max-Tucker rank naturally arises from applications of the Tucker decomposition when people assume that for and fix the value of [3, 12]. Then this means that tensors of max-Tucker ranks not greater than are used. In the following, we introduce a new tensor rank function, which is also associated with the Tucker decomposition, but is different from the max-Tucker rank. We may replace (3.4) by

 r=submax{r1,⋯,rm}. (3.5)

Then we have the following theorem.

###### Theorem 3.4

The function defined by (3.5) is a strongly proper tensor rank function. But it is not subadditive.

Proof We first show that function defined by (3.5) is a tensor rank function. It suffices to show that Property 1-6 are all satisfied.

1) Suppose that for is a zero tensor. Then are zero matrices for all . This implies that , which means that . By (3.5), we have . On the other hand, assume that for some with . This means that for some , , and hence , . Therefore, if and only if

Suppose that , then is not zero tensor and hence then for all . Since is the submax-Tucker rank, there exists such that . Without loss of generality, we assume that for . Let . Similar to discussion in proof of Theorem 3.3, is rank one for all . Thus

 X=a(1,p)∘⋯∘(a(m,1)+λ2a(m,2)+λ3a(m,3)+⋯+λ¯ra(m,¯r)),

for some . Clearly, such is a rank one tensor.

Conversely, if is a rank one tensor, then for some nonzero vectors . Then and for all . Thus .

Based on the above analysis, Property 1 is satisfied.

2) Denote Then is an identity matrix for , and hence . Thus, .

3) When , then , and for any . Clearly, and when . Hence is the same as matrix rank.

4) Suppose that . For any , and any , and hence . Hence .

5) Suppose that and is any permuted tensor of . Then will be for . So . Hence and the result holds.

6) Suppose that is a subtensor of . Then for all , will be a submatrix of and since is matrix rank. So . Suppose that is an essential subtensor of , then is an essential subtensor of and hence . So we can assert that .

Now we conclude that defined by (3.4) is a tensor rank function.

It is clear that such a tensor rank function is proper from its definition. Furthermore, we have that such rank is also subadditive since matrix rank is subadditive.

In addition, we consider with where is the identity matrix of three dimension. Hence . Hence we conclude that such a tensor rank function is not strongly proper.

The strongly proper property of such a tensor rank function is clear and hence it suffices to show that it is not subadditive.

Let with and

 Yijk=0  if  i>n1, j>n2, k>n3,Zpqs=0  if  p≤n1, q≤n2,s≤n3.

It is assumed that , and , . Then for and . So since and . Therefore, we conclude that such a tensor rank function is not subadditive.

.

Thus, we call this tensor rank function the submax-Tucker rank in this paper, and denote it as submax-TucRank.

###### Proposition 3.5

We have submax-TucRank max-TucRank and submax-TucRank max-TucRank. Thus, max-TucRank .

Proof Clearly, mid-TucRank max-TucRank. To see mid-TucRank max-TucRank, we consider the following counterexample.

Consider the tensor with its nonzeros entries . By observation, we have that and , which implies that submax-TucRankmax-TucRank and the result is arrived here.

.

We cannot replace submax in (3.5) by the third smallest value in , as this will violate Property 1 of Definition 2.1.

## 4 The Closure of a Strongly Proper Tensor Rank Function

We know that the submax-Tucker rank function is a strongly proper tensor rank functions. Based upon this, we may define the closure of a strongly proper tensor rank function.

We may have the following definitions.

###### Definition 4.1

Suppose that is a strongly proper tensor rank function. Let with , and . If we have

 r(X)=submax{n1,⋯,nm}, (4.6)

then we say that is of full rank.

###### Definition 4.2

Suppose that is a strongly proper tensor rank function. We say that is closed if for any with , and , either is of full rank, or has a subtensor such that and is of full rank. In particular, zero tensors and rank one tensors are regarded as full of rank.

These two definitions lead to some questions. First, is the submax-Tucker rank function closed? If this can be proved, then the next question is: Are all tensor rank functions closed? If not, can we give a counterexample? We may also have the following definition.

###### Definition 4.3

Suppose that is a strongly proper tensor rank function. We may define as the closure of by

 ¯r(X)=max{r(Y):Y is a subtensor of X, and of full r rank},

for any