# relationship between $\mathbb{R}^2 \otimes \mathbb{R}^2 \ \text{and}\ \mathbb{R}^4 ?$

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 $\mathbb{R}^2 \otimes \mathbb{R}^2 \ \text{and}\ \mathbb{R}^4 ?$

• $\mathbb{R}^2 \otimes \mathbb{R}^2$ 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. – Adayah Aug 18 '17 at 14:10
• It seems that i commit a mistake on the definition of tensor product between sets, thanks for your correction! – 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 $\mathbb R^2\otimes \mathbb R^2\cong \mathbb R^4$.