I hope you can help me with this:

Show that $\Bbb{R}^2 \times \Bbb{R}^2$ and $\Bbb{R}^2 \otimes \Bbb{R}^2$ are isomorphic, and specify an isomorphism.


  • $\begingroup$ I assume you mean $\mathbb{R^2}$ written \mathbb{R^2}. What have you tried so far? $\endgroup$ – Zach Stone Jun 18 '15 at 22:30
  • $\begingroup$ Please format your question in TeX notation, provide some context for the question, and indicate what you've tried so far. Just dumping a poorly-formatted question here without any explanation or effort of your own is not going to be productive. $\endgroup$ – anomaly Jun 18 '15 at 22:33
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    $\begingroup$ Be gentle! Someone asking how to show that $2 + 2 = 2 \times 2$ should surely be treated with a little kindness on MSE, don't you think? $\ddot{\smile}$. $\endgroup$ – Rob Arthan Jun 18 '15 at 23:24

In the beginning I just specify what Rob Arthan meant. There are following three theorems. For simplicity considered vector spaces are over filed $\mathbb{R}.$

1.If $V$ and $W$ are finite dimensional vector spaces then $\dim(V\times W)=\dim(V)+\dim(W).$

Be awere that in category of vector spaces we can use $\times$ and $\oplus$ interchangeably.

2.If $V$ and $W$ are finite dimensional vector spaces then $\dim(V\otimes W)=\dim(V)\cdot\dim(W).$

The third one

3.If $V$ and $W$ are finite dimensional vector spaces such that $\dim(V)=\dim(W)$ then $V$ and $W$ are isomorphic.

Using these three theorems to your case you see that $\Bbb{R}^2 \times \Bbb{R}^2$ and $\Bbb{R}^2 \otimes \Bbb{R}^2$ are isomorphic because $$2+2=2\cdot 2.$$ But to construct this isomorphism you have to refer to basis in $\Bbb{R}^2 \times\Bbb{R}^2$ and $\Bbb{R}^2 \otimes \Bbb{R}^2.$ So let $e_1=(1,0)$ and $e_2=(0,1).$ Hence $$\{(e_1,0),(e_2,0),(0,e_1),(0,e_2)\}$$ is a base in $\Bbb{R}^2 \times\Bbb{R}^2$ and $$\{e_1\otimes e_1,e_1\otimes e_2,e_2\otimes e_1,e_2\otimes e_2\}$$ is a base in $\Bbb{R}^2 \otimes \Bbb{R}^2.$

Now the required isomorphism is for example $\mathbb{R}$-linear map $\phi:\Bbb{R}^2 \times\Bbb{R}^2\rightarrow \Bbb{R}^2 \otimes \Bbb{R}^2$ such that$$\phi:(e_1,0)\mapsto e_1\otimes e_1,(e_2,0)\mapsto e_2\otimes e_1,(0,e_1)\mapsto e_1\otimes e_2,(0,e_2)\mapsto e_2\otimes e_2.$$ $\phi$ is well defined, cause to define a linear map it is sufficient to determine how it acts on a basis. I left you checking that it is in fact an isomorphism.

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  • $\begingroup$ Strictly speaking you also want to require that all vector spaces are over the same underlying field. $\endgroup$ – Klaus Draeger Jun 19 '15 at 12:43
  • $\begingroup$ @KlausDraeger You are right. I will edit the answer. $\endgroup$ – Fallen Apart Jun 19 '15 at 12:44
  • $\begingroup$ @FallenApart It's OK ! Don't write it dirty $\endgroup$ – user18537 Jun 19 '15 at 12:46

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