# A tensor over $\mathbb R$ is sum of two decomposable tensors

Let $$W$$ be a vector space over $$\mathbb R$$ with basis $$\{a,b\}$$. Consider the tensor $$a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b.$$

a) Show that this tensor cannot be presented as sum of two decomposable tensors.

b) Show that such presentation is possible in complexification of $$W$$.

My only attempt was to write an equation $$(\alpha a + \beta b)\otimes (\gamma a + \delta b) \otimes (\epsilon a + \zeta b) + (\alpha' a + \beta' b)\otimes (\gamma' a + \delta' b) \otimes (\epsilon' a + \zeta' b) =$$ $$= a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b,$$ and from equality of coefficients deduce a system of 8 equalities and 12 variables. I think I can brute force it some way, but I was wondering if there are any other approaches.

• I would start by b), and try something like this first $(a+bi)\otimes(a-bi)=a\otimes a -i\,a\otimes b+i\,b\otimes a+b\otimes b$, it should be something along these lines. Nov 28, 2020 at 21:31

I'll start by stating some common theorems with tensor products.

## 1. Preliminaries

Theorem 1 (Zero-sum)
Let $$U$$,$$V$$ be two vector spaces on $$F$$, if for $$n$$ linearly independent vectors $$x_1,x_2,...,x_n \in U$$, there are $$n$$ vectors $$v_1,...,v_n \in V$$ such that: $$x_1 \otimes v_1 +x_2 \otimes v_2+...+x_n \otimes v_n= 0$$ then $$v_1=v_2=...=v_n=0$$ $$\square$$

Theorem 2 (Span)
Let $$U$$,$$V$$ be two vector spaces on $$F$$. $$W$$ is a subspace of $$V$$.
Let $$t$$ be a tensor in $$U \otimes W$$ and if for $$n$$ linearly independent vectors $$x_1,x_2,...,x_n \in U$$, there are $$n$$ vectors $$v_1,...,v_n \in V$$ such that: $$t=x_1 \otimes v_1 +x_2 \otimes v_2+...+x_n \otimes v_n$$ then $$\text{span}(v_1,...,v_n) \subset W$$
$$\square$$
Remark 3: We will use mainly theorem 2, the proof of theorem 2 is based on theorem 1. Furthermore, we only need this theorem in the special case where $$\dim W=2$$

Assume there are vectors $$u_1 , u_2, u_3,v_1,v_2,v_3 \in W$$ such that: $$a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b= u_1\otimes u_2 \otimes u_3+u_1\otimes u_2 \otimes u_3 =:t$$ On one hand, we see that: $$t \in W\otimes( \underbrace{\text{span}( u_2\otimes u_3, v_2 \otimes v_3)}_{=:U})$$ On the other hand, $$t= a \otimes ( a \otimes a +b \otimes b)+ b\otimes ( -b \otimes a + a \otimes b)$$ Then by theorem 2, we imply that $$a \otimes a +b \otimes b$$ and $$-b \otimes a + a \otimes b$$ lie in $$U$$.
By theorem 1, none of two above vector is null (and clearly they are independent) , whereas $$\dim(U)\le 2$$.
Therefore, $$\dim(U)=2$$ and $$U = \text{span}( a \otimes a +b \otimes b , -b \otimes a + a \otimes b )$$

So there are two elements $$p,q \in F$$ ( $$F$$ is the field on which we define $$W$$, in this case it is either $$\mathbb{R}$$ or $$\mathbb{C}$$) such that: $$p (a \otimes a +b \otimes b ) +q( -b \otimes a + a \otimes b ) =u_2 \otimes u_3$$

Or , $$a \otimes( pa +qb) +b \otimes ( bp-qa) = u_2 \otimes u_3$$ Again, by theorem $$2$$ ( or by any other straightforward reasoning), as $$a,b$$ are independent, $$pa+qb$$ and $$bp-qa$$ cannot be independent ( as $$\dim( \text{span}(u_3)) = 1$$)
As $$a,b$$ are independent, the two above vector are linearly dependent iff $$p^2+q^2=0$$ This cannot occur in $$\mathbb{R}$$ unless $$p=q=0$$ which also leads to a contradiction.

Hence the conclusion for $$\mathbb{R}$$.

The conclusion for $$\mathbb{C}$$ (if true) is just a result from some simple calculations.

Remark Indeed, many arguments I presented above can be omitted or shortened given that we are familiar with the tensor product. I just tried to show this point as clear as possible.

• Tip: Use \dim to display it more nicely as $\dim$. You can also use \text{} or \operatorname{} to display $\text{span}$ and $\operatorname{span}$ respectively. Dec 3, 2020 at 1:24
• @K.defaoite Thank you for your advice, I'll note that. Dec 3, 2020 at 1:27
• "By theorem 1, none of the above vector is null." Is theorem 1 needed here? I thought $a\otimes a + b\otimes b$ and $-b \otimes a + a \otimes b$ are non-null as non-trivial combinations of basis vectors in $W \otimes W$
– dnes
Dec 8, 2020 at 7:00
• @dnes: Your remark is right. It's my habit, I just don't like using "basis" in my argument. Dec 8, 2020 at 14:55

Here is a solution that reduces the problem to linear algebra. (There are probably other solutions.)

### Reduction to linear algebra

Let $$\{\varphi,\chi\}$$ be a basis of $$W^*$$ with $$\varphi(a) = \chi(b) = 1$$ and $$\varphi(b) = \chi(a) = 0$$. Suppose that $$a \mathbin{\otimes} a \mathbin{\otimes} a - b \mathbin{\otimes} b \mathbin{\otimes} a + a \mathbin{\otimes} b \mathbin{\otimes} b + b \mathbin{\otimes} a \mathbin{\otimes} b = u \mathbin{\otimes} v \mathbin{\otimes} w + x \mathbin{\otimes} y \mathbin{\otimes} z.$$ Applying $$\varphi$$ and $$\chi$$ in the first coordinate, we find \begin{align*} a \mathbin{\otimes} a + b \mathbin{\otimes} b &= \varphi(u) v \mathbin{\otimes} w + \varphi(x) y \mathbin{\otimes} z;\\[1ex] -b \mathbin{\otimes} a + a \mathbin{\otimes} b &= \chi(u) v \mathbin{\otimes} w + \chi(x) y \mathbin{\otimes} z.\tag*{(1)} \end{align*} Therefore $$C := v \mathbin{\otimes} w$$ and $$D := y \mathbin{\otimes} z$$ are rank one matrices in $$W \mathbin{\otimes} W$$ with the property that both $$a \mathbin{\otimes} a + b \mathbin{\otimes} b$$ and $$-b \mathbin{\otimes} a + a \mathbin{\otimes} b$$ are linear combinations of $$C$$ and $$D$$. We will show that this is impossible over $$\mathbb{R}$$ and possible over $$\mathbb{C}$$. In the latter case, the obtained matrices $$C$$ and $$D$$ can easily be used to solve the original problem.

### Solution

The preceding remarks show that the problem is equivalent to the following: find $$2 \times 2$$ matrices $$C = (c_{ij})$$ and $$D = (d_{ij})$$, each of rank one, and scalars $$\lambda_1,\lambda_2,\mu_1,\mu_2$$ such that $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \lambda_1 C + \lambda_2 D \qquad \text{and} \qquad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \mu_1 C + \mu_2 D. \tag*{(2)}$$ We prove that this is impossible over $$\mathbb{R}$$. Since $$C$$ and $$D$$ have rank one, we must have $$\lambda_1,\lambda_2,\mu_1,\mu_2 \neq 0$$, and we may choose some non-zero $$p \in \ker(C)$$. But then it follows from $$(2)$$ that \begin{align*} p &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}p = (\lambda_1 C + \lambda_2 D)p = \lambda_2 Dp;\\[1ex] p' &:= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}p = (\mu_1 C + \mu_2 D)p = \mu_2 Dp; \end{align*} where $$p'$$ is $$p$$ rotated by $$-90^\circ$$. But then $$Dp$$ is simultaneously parallel ($$Dp = \frac{1}{\lambda_2} p \neq 0$$) and perpendicular ($$Dp = \frac{1}{\mu_2} p' \neq 0$$) to $$p$$, with $$Dp \neq 0$$. This is impossible.

Finally, over $$\mathbb{C}$$ one may take $$C = \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \qquad \text{and} \qquad D = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}.$$ In order to turn this into a solution of the original problem, choose $$u,x \in W$$ such that $$\varphi(u),\varphi(x),\chi(u),\chi(x)$$ are the desired scalars in equation $$(1)$$.

### Closing remarks

It can be shown that the approach taken here always works to determine the rank of a $$3$$-tensor:

Proposition ([Lan12, Thm 3.1.1.1]). Let $$T \in U \mathbin{\otimes} V \mathbin{\otimes} W$$. Then $$\text{rank}(T)$$ equals the number of rank one matrices needed to span $$T(U^*) \subseteq V \mathbin{\otimes} W$$.

Here we interpret $$T$$ as a linear map $$U^* \to V \mathbin{\otimes} W$$ in the natural way: if $$T = \sum_{i=1}^r u_i \mathbin{\otimes} v_i \mathbin{\otimes} w_i$$ and $$\psi \in U^*$$, then $$T(\psi) = \sum_{i=1}^r \psi(u_i) v_i \mathbin{\otimes} w_i$$. The proposition is not hard to prove, and generalizes the "reduction to linear algebra" paragraph above to arbitrary $$3$$-tensors.

### References.

[Lan12] J. M. Landsberg, Tensors: Geometry and Applications, Graduate Studies in Mathematics 128, American Mathematical Society, 2012.