Tensor Product: Vector Spaces

Question

Reference

Foundation for: Hilbert Spaces: Tensor Product

Problem

Given a vector spaces $V$ and $W$.

Take its algebraic tensor product: $\tau:V\times W\to V\otimes W$

How to prove that the image is proper in general: $$\mathrm{im}\tau=V\otimes W\iff\dim V\leq1\lor\dim W\leq1$$

(Presuming only the universal property!)

My attempt was to consider linearly independent vectors: $v\neq\kappa v',\,w\neq\lambda w'$

For contradiction assuming it holds: $v\otimes w+v'\otimes w'=v_0\otimes w_0$

But from here I don't know how to proceed.

Are you allowed to know anything about a basis for the tensor product? — Thomas, Dec 19 '14 at 14:13
@Omnomnomnom: My apologiesfor rolling back your edit. I prefer the symbol, here. — C-star-W-star, Dec 19 '14 at 14:25
In some fields $a \lor b = \max{a,b}$, which makes reading the question as written confusing. — Ben Grossmann, Dec 19 '14 at 14:28
@Omnomnomnom: In any field the wedge denotes an operator with certain universal properties; the logical or, the set union and the order maximum being some of them. ;) — C-star-W-star, Dec 19 '14 at 14:31
@Daniel: Ah ok so the simple tensors and higher rank tensors. Do you have nice a link forthis? — C-star-W-star, Dec 19 '14 at 14:36
@Thomas: Yep, so by the universal property it induces a basis: $e_k\otimes f_l$ — C-star-W-star, Dec 19 '14 at 16:10
@Freeze_S See the proposition used in this answer: http://math.stackexchange.com/questions/981056/rank-of-tensors-in-terms-of-ranks-of-associated-linear-maps/981511#981511 The link for the paper is there too. — Daniel, Dec 19 '14 at 21:09

score 3 · Accepted Answer · answered Dec 19 '14 at 15:03

For simplicity assume that $V$ and $W$ both have dimension $2$ and have bases $\{e_1, e_2\}$ and $\{f_1, f_2\}$ respectively.

I assume that $\tau$ is defined by $$ \tau (a_1e_1 + a_2e_2, b_1f_1 + b_2f_2) = a_1b_1(e_1\otimes f_1) + a_2b_1(e_2\otimes f_1) + a_1b_2(e_1\otimes f_2) + a_2b_2(e_2\otimes f_2). $$

Now say that $\tau$ surjective, then there you would have a "solution" to $$ \tau (a_1e_1 + a_2e_2, b_1f_1 + b_2f_2) = e_1\otimes f_1 + e_2\otimes f_2. $$ From above this would give you equations $$\begin{align} a_1b_1 &= 1 \\ a_2b_1 &= 0 \\ a_1b_2 &= 0 \\ a_2b_2 &= 1. \end{align} $$ But this system doesn't have any solutions. So you have a contradiction.

(You can make this more general.)

Nice answer. Thanks. =D (How do you mean more general?) – C-star-W-star Dec 19 '14 at 16:00 — C-star-W-star, Dec 19 '14 at 16:00
Besides, do you have an idea for the linked thread? – C-star-W-star Dec 19 '14 at 16:12 — C-star-W-star, Dec 19 '14 at 16:12

Tensor Product: Vector Spaces

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