Tensor Product of Spaces

[klackity]

Tell me if this is true:

We are given vector spaces V₁, V₂, ..., V_n of dimensions d₁, d₂, ..., d_n respectively.

Let V = V₁ $$\otimes$$ V₂ $$\otimes$$ ... $$\otimes$$ V_n

Claim: Any element v $$\in$$ V can be represented in the following form:

$$\sum$$_i=1...R (v_1,i $$\otimes$$ ... $$\otimes$$ v_n,i)

Where R = - MAX {d_j} + $$\sum$$_j=1..n d_j

And where v_j,i $$\in$$ V_j.

In other words, there is an upper bound of R on the number of "elementary" tensors w_i needed needed to represent a particular tensor v in V (where an "elementary" tensor w_i is one which can be written in the form w_i = v₁ $$\otimes$$ v₂ $$\otimes$$ ... v_n, where v_j $$\in$$ V_j). The upper bound R is simply the sum of all the d_j except for the d_j0 which is maximal.

Is this true?

 

[micromass]

Well, you know that the dimension of V is $$d_1d_2...d_n$$. So I guess that you probably need this many vectors to express an arbitrary vector in V...

 

[klackity]

Well, you know that the dimension of V is $$d_1d_2...d_n$$. So I guess that you probably need this many vectors to express an arbitrary vector in V...

Well, yes, you need d₁*d₂*...*d_n vectors to form a basis for V.

But I am asking something slightly different. I'll give a concrete example:

Suppose n=2 and V₁ = R² and V₂ = R₃^* (the space of linear functionals on R³)

Then V = V₁ $$\otimes$$ V₂ $$\cong$$ R⁶.

Any tensor in V is a 2x3 matrix:

M = $$\left[ \begin{array}{cccc} a & b & c \\ d & e & f \end{array} \right]$$

Now, the problem is that not every such matrix is the tensor product of two elements v₁ and v₂ of V₁ and V₂ respectively. Any maximal-rank matrix is able to be represented as the sum of exactly two (and no fewer) such "elementary" tensors (where an "elementary" tensor is the tensor product of two elements v₁ and v₂ (need not be basis vectors) of V₁ and V₂ respectively). A non-maximal rank matrix (i.e., one with rank strictly less than 2 = min{2,3}) can be represented as a sum of fewer "elementary" tensors (e.g., one).

V contains elements M in the correct form such that I can put a vector x (a column vector) to the right of M, and I will get another vector back as a result of that "multiplication". If I put a linear functional (row vector) to the left of it, I will get back another linear functional (row vector). If I put both a vector to the right, and a linear functional to the left, I will get back a real number.

The matrix M itself can be thought of as two linear functionals on R³ (row vectors) attached to two particular vectors (column vectors) in R². The topmost row of M is a linear functional attached to the vector (1,0) (in column form), and the bottommost row of M is a linear functional attached to the vector (0,1) (in column form). When you put a vector (in R³) to the right of M, you are using those two linear functionals to determine coefficients a₁, a₂ (for the top and bottom rows, repsectively) which will factor into the sum a₁(1,0) + a₂(0,1).

You can also paint a similar picture for thinking about multiplying a linear functional (row vector) on the left of M. In this case, M can be thought of as consisting of three vectors (in R²) each attached to one of the row vectors (1,0,0), (0,1,0), and (0,0,1). However, since there are three such linear functionals, they cannot be linearly independent. So you really only need two linear functionals attached (repsectively) to two row vectors, although this time the row vectors may not be basis vectors.

So either way you think about it, you see that you really only need two bits of "elementary" information to represent any particular matrix M in V, even though there are 6 basis elements to V.

The "elementary" bits of information can be represented as a pair (x,x') in V₁xV₂ (modulo an equivalence relation).

For matrices, the number of "elementary" bits of information you need is just the maximal number of linearly independent rows (or columns) of M. For higher order tensors, I postulate that it is something a bit more complicated, but that it has an upper bound (which is achieved) of R (where R is defined above).

 

[mathwonk]

well it looks true for n=2. can you use induction?

 

[blue_raver22]

Yes, this statement is true. The tensor product of vector spaces V1, V2, ..., Vn is defined as the set of all linear combinations of elementary tensors v1 \otimes v2 \otimes ... \otimes vn, where vj \in Vj. The dimension of this tensor product space is given by the product of the dimensions of the individual vector spaces, i.e. dim(V) = d1 * d2 * ... * dn. Therefore, any element v \in V can be represented as a linear combination of elementary tensors, where the number of elementary tensors needed is at most equal to the dimension of V.

The claim given in the statement is stating that there is an upper bound on the number of elementary tensors needed to represent a particular tensor v in V. This upper bound is equal to the sum of all the dimensions of the individual vector spaces, except for the maximum dimension. This makes sense because the maximum dimension will already be represented in the other dimensions, so it does not need to be counted again.

In summary, the statement is true and provides a useful understanding of the properties of tensor products of vector spaces. It highlights the structured and systematic nature of tensor products, and the relationship between the dimensions of the individual vector spaces and the resulting tensor product space.