# Tensor Product Vector Space Example

While trying to study representation theory, I came across the tensor product of vector spaces - a new topic to me (although done some multinear algebra in the past). A definition I came across of the tensor product of vector spaces $U$ and $V$ is that it is the vector space $U\otimes{V}$ such that bilinear maps from $U\times{V}$ into a vector space W, are linear maps from $U\otimes{V}$ onto W. In this way, $U\otimes{V}$ reduces multilinear algebra to the more simple linear algebra.

I am trying currently to now find an example in understanding this topic. I considered letting $U=\mathbb{R^3}$, $V=\mathbb{R^3}$, $W=\mathbb{R^3}$, and then considering the bilinear map that is the cross product from $U\times{V}$ into W. However, I can't seem to construct the space $U\otimes{V}$ such that the cross product is a linear map from $U\otimes{V}$ into $W$. Could someone help me with identifying this space/supplement with further examples of the tensor product of vector spaces. Thanks.

• $U\otimes V$ is nine-dimensional; if we let $e_1$, $e_2$, $e_3$ be basis vectors of $U=V$, then one basis for $U\otimes V$ is $e_1\otimes e_1$, $e_1\otimes e_2,\ldots,e_3\otimes e_3$. May 2, 2018 at 19:33
• @LordSharktheUnknown I understand that the elements of the tensor product are linear combinations of the tensor products of the basis vectors, but how does one (for example in the cross product case) actually compute $e_1\otimes{e_1}$,etc. May 2, 2018 at 19:42
• @IskyMathews ... May 2, 2018 at 19:43

Let us consider the basis of $U \otimes V$. As we are dealing with $\mathbb{R}^3 \otimes \mathbb{R}^3$, the basis $B$ is the set of all $e_i \otimes e_j$ for $i,j \in \{1, 2, 3\}$. Note that for the cross product, the multiplication table is as follows

$$\begin{array}{c|c|c|c|} \times & e_1 & e_2 & e_3 \\ \hline e_1 & 0 & e_3& -e_2\\ \hline e_2 & -e_3 & 0 & e_1 \\ \hline e_3 & e_2 & -e_1 & 0 \\ \hline \end{array}$$

Then define a map $W: B \to \mathbb{R}^3$ by the following multiplication table $$\begin{array}{c|c|c|c|} W & e_1 & e_2 & e_3 \\ \hline e_1 & 0 & e_3& -e_2\\ \hline e_2 & -e_3 & 0 & e_1 \\ \hline e_3 & e_2 & -e_1 & 0 \\ \hline \end{array}$$

As an example of how the table is to be read, this table says that $W(e_1 \otimes e_2) = e_3$.

Now that we have defined $W$ on the basis, we can extend to the rest of $\mathbb{R}^3 \otimes \mathbb{R}^3$. This is because the tensor product and the cross product have the same structure. The cross product is a bilinear map, that is:

\begin{align} (v_1 + v_2) \times w = v_1 \times w + v_2 \times w \\ v \times (w_1 + w_2) = v \times w_1 + v \times w_2 \\ \end{align}

and the tensor space follows a set of "bilinear" axioms:

\begin{align} (v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w \\ v \otimes (w_1 + w_2) = v \otimes w_1 + v \otimes w_2 \\ \end{align}

These are very similar. Now, extending $W$ linearly, we truly see that $$W(v \otimes w) = v \times w$$ As an example, which illuminates the proof of this fact, consider the following computation:

\begin{align} W((1\mathbf{e}_1 + 2\mathbf{e}_3) \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3)) &= \\ W(1\mathbf{e}_1 \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3) + 2\mathbf{e}_3 \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3)) &= \\ W(1\mathbf{e}_1 \otimes 3\mathbf{e}_2 + 1\mathbf{e}_1 \otimes 4\mathbf{e}_3 + 2\mathbf{e}_3 \otimes 3\mathbf{e}_2 + 2\mathbf{e}_3 \otimes 4\mathbf{e}_3) &= \\ W(1\mathbf{e}_1 \otimes 3\mathbf{e}_2) + W(1\mathbf{e}_1 \otimes 4\mathbf{e}_3) + W(2\mathbf{e}_3 \otimes 3\mathbf{e}_2) + W(2\mathbf{e}_3 \otimes 4\mathbf{e}_3) &= \\ W(3(\mathbf{e}_1 \otimes \mathbf{e}_2)) + W(4(\mathbf{e}_1 \otimes \mathbf{e}_3)) + W(6(\mathbf{e}_3 \otimes \mathbf{e}_2)) + W(8(\mathbf{e}_3 \otimes \mathbf{e}_3)) &= \\ 3W(\mathbf{e}_1 \otimes \mathbf{e}_2) + 4W(\mathbf{e}_1 \otimes \mathbf{e}_3) + 6W(\mathbf{e}_3 \otimes \mathbf{e}_2) + 8W(\mathbf{e}_3 \otimes \mathbf{e}_3) &= \\ 3\mathbf{e}_3 - 4\mathbf{e}_2 - 6\mathbf{e}_1 + 8\cdot\mathbf{0} &= \\ -6\mathbf{e}_1 - 4\mathbf{e}_2 + 3\mathbf{e}_3 \end{align} which is indeed the cross product of those two vectors. Steps 4 and 6 rely on the fact $W$ had been extended linearly, while the rest of the steps used the bilinear properties of the tensor product. (Compare this to explicitly finding the cross product by expanding and multiplying!)

Conclusion: The tensor product allows us to use its axioms of bilinearity to "mimic" the properties of a bilinear object or map.

• @Jerry Opoku-Mensah thank you very much for your answer, however, in the second line it seems to my unknowing-mind that you have defined the tensor product on itself... What is $e_i\otimes{e_j}$ and what would be the computation for it... I still do not understand what $\mathbb{R^3}\otimes{\mathbb{R^3}}$ 'looks like'... May 2, 2018 at 22:56
• The tensor product is defined on itself because you defined it on itself, $\mathbb{R}^3 \otimes \mathbb{R}^3$ is a self-product. I will post another answer describing what it "looks like," if that troubles you.
– user211599
May 2, 2018 at 23:26
• Apologies if I still dont understand, but I can't quite understand what you mean by its defined on itself - i.e. what is $e_1\otimes{e_2}$ in the standard basis for $\mathbb{R^3}$. I get that U tensor V is the space of all linear combinations on tensor products of the basis, but how does one compute the tensor product of two basis vectors May 3, 2018 at 13:02
• You don't "compute" them. They are new objects. That is like asking: "I get that 1 + 2i is the sum of a real and imaginary number, but how does one compute 1 + 2i?"
– user211599
May 3, 2018 at 20:22

Let $U$ be the space of real polynomials in the "formal variable" $x$ with degree $\leq 3$. $U$ is an $\mathbb{R}$-vector space, as you can easily verify. What does the canonical basis of $U$ look like? I bet you already know that it is $$B_U =\{x^0, x^1, x^2, x^3\}$$ Here is an example element of $U$: $$3x^3 + x^2 + 10$$

Let's also consider the vector space $V$ of real polynomials in the "formal variable" $y$ with degree $\leq 2$. This time, the basis is $$B_V = \{y^0, y^1, y^2\}$$ Here is an example element of $V$: $$11y^2 - 9$$

Note that we are only consider the "form" of the variable, in this sense, because $x^0$ and $y^0$ are different objects, they are not considered the same.

Now by definition, here is the basis of $B_U \otimes B_V$. $$B_{U \otimes V} = \begin{Bmatrix} x^0 \otimes y^0 & x^1 \otimes y^0 & x^2 \otimes y^0 & x^3 \otimes y^0 \\ x^0 \otimes y^1 & x^1 \otimes y^1 & x^2 \otimes y^1 & x^3 \otimes y^1 \\ x^0 \otimes y^2 & x^1 \otimes y^2 & x^2 \otimes y^2 & x^3 \otimes y^2 \\ \end{Bmatrix}$$ Using this basis, here is an element of $U \otimes V$. $$5(x^3 \otimes y^2) + 4(x^3 \otimes y^1) - 10(x \otimes y^2) + 6(x^0 \otimes y^1) - 10(x^0 \otimes y^0)$$ See how this looks like a polynomial in $x$ and $y$? The tensor product space acts exactly in the way you'd think, using bilinearity(acting as distributivity) to allow us to "multiply" the elements of two vector spaces. Using the examples from before,

$$(3x^3 + x^2 + 10) \otimes (11y^2 - 9) = \\ 33(x^3 \otimes y^2) + 11(x^2 \otimes y^2) + 110(x^0 \otimes y^2) - 27(x^3 \otimes y^0) - 9(x^2 \otimes y^0) - 90(x^0 \otimes y^0)$$

Which looks like the actual product

$$(3x^3 + x^2 + 10)(11y^2 - 9) = \\ 33x^3y^2 + 11x^2y^2 + 110y^2 - 27x^3 - 9x^2 - 90$$

Conclusion: The tensor product space is a space where the elements from each original space may be multiplied. This multiplication is distributive and allows different basis elements to interact, meaning it looks a lot like polynomial multiplication.

Note: Not every element in $U \otimes V$ is a multiplication of elements in $U$ and $V$. The example element of $U \otimes V$ I gave before is not factorable, as an example. Look here for more information.