when i consider tensor product on standard orthonormal basis of $\mathbb{R}^2$, for example $$ \left( \begin{array} {c}1 \\ 0 \end{array}\right) \otimes \left( \begin{array} {c}1 \\ 0 \end{array}\right)=\left( \begin{array} {c}1 \\ 0\\0\\0 \end{array}\right), $$ $$ \left( \begin{array} {c}1 \\ 0 \end{array}\right) \otimes \left( \begin{array} {c}0 \\ 1 \end{array}\right)=\left( \begin{array} {c}0 \\ 1\\0\\0 \end{array}\right), $$ $$ \left( \begin{array} {c}0 \\ 1 \end{array}\right) \otimes \left( \begin{array} {c}1 \\ 0 \end{array}\right)=\left( \begin{array} {c}0 \\ 0\\1\\0 \end{array}\right), $$ $$ \left( \begin{array} {c}0 \\ 1 \end{array}\right) \otimes \left( \begin{array} {c}0 \\ 1 \end{array}\right)=\left( \begin{array} {c}0 \\ 0\\0\\1 \end{array}\right), $$ hence i feel $\mathbb{R}^2 \otimes \mathbb{R}^2 =\mathbb{R}^4 ? $ But when i consider $\mathbb{R}^2 \otimes \mathbb{R}^2 $ in the following way: for any $ \left( \begin{array} {c}a_1 \\ b_1 \end{array}\right), \left( \begin{array} {c}a_2 \\ b_2 \end{array}\right) \in \mathbb{R}^2 $ $$ \left( \begin{array} {c}a_1 \\ b_1 \end{array}\right) \otimes \left( \begin{array} {c}a_2 \\ b_2 \end{array}\right)=\left( \begin{array} {c}a_1 a_2 \\ a_1 b_2\\b_1 a_2\\b_1 b_2 \end{array}\right) $$ $\mathbb{R}^4 $ consists of all elements like $$\left( \begin{array} {c}x_1 \\ x_2\\x_3\\x_4 \end{array}\right)=\left( \begin{array} {c}a_1 a_2 \\ a_1 b_2\\b_1 a_2\\b_1 b_2 \end{array}\right) a_i, b_i \in \mathbb{R},\ \ i=1,2 $$ without considering zero-case for $ a_i, b_i,\ \ i=1,2 $, we can see that $ \frac{x_1}{x_3}=\frac{x_2}{x_4}$ which means $ x_1,x_2,x_3,x_4 $ can't taking value independently,then i would say $\mathbb{R}^2 \otimes \mathbb{R}^2 \neq \mathbb{R}^4 $.

Now, i obtain contradiction, who can tell me what's wrong, and the true answer on if $\mathbb{R}^2 \otimes \mathbb{R}^2= \mathbb{R}^4? $

I trend to think it is incorrect.

And if $\mathbb{R}^2 \otimes \mathbb{R}^2 \neq \mathbb{R}^4 $, then what's the relationship between $\mathbb{R}^2 \otimes \mathbb{R}^2 \ \text{and}\ \mathbb{R}^4 ?$

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    $\begingroup$ $\mathbb{R}^2 \otimes \mathbb{R}^2$ consists of linear combinations of elements $u \otimes v$ where $u, v \in \mathbb{R}^2$. But not every element is of the form $u \otimes v$, e.g. $\binom{1}{0} \otimes \binom{1}{0} + \binom{0}{1} \otimes \binom{0}{1}$. So $\mathbb{R}^2 \otimes \mathbb{R}^2 \cong \mathbb{R}^4$ is correct. See here. $\endgroup$ – Adayah Aug 18 '17 at 14:10
  • $\begingroup$ It seems that i commit a mistake on the definition of tensor product between sets, thanks for your correction! $\endgroup$ – Yidong Luo Aug 18 '17 at 14:30

When $V$ and $W$ are finite dimensional spaces, it is well known that $$\dim(V\otimes W) = \dim V\cdot \dim W.$$

Also, $\dim V = n$ if and only if $V\cong \mathbb R^n$.

Combining the two, we have $\dim(\mathbb R^2\otimes \mathbb R^2) = (\dim \mathbb R^2)^2 = 4 = \dim \mathbb R^4$ and hence $\mathbb R^2\otimes \mathbb R^2\cong \mathbb R^4$.


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