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I saw here that to prove two infinite dimensional vector spaces not being isomorphic involves their dimensions (if they have the same dimension, then they are isomorphic). However, I have only been given this following theorem to work with:

Let $V$ and $W$ be two finite dimensional vector spaces (over the same field of scalars). Then $V$ is isomorphic to $W$ if and only if $dimV=dimW$.

I have been asked to prove the following vector spaces $V,W$ isomorphic or not. $V = \{A\in \mathcal{M}_{22} \ | \ tr(A) = 0 \}, W = \mathbb{R}^2$.

I suspect that they are not isomorphic, since $\mathcal{M}_{22}$ has dimension 4 and $\mathbb{R}^2$ has dimension 2. My question is: would the method of proving that if two infinite dimensional vector spaces have different dimensions then they are not isomorphic be the best way to prove this? Or is there an easier way to do it?

user10448599
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  • Both $V$ and $W$ are finite dimensional, so I don't see why you're concerned about infinite-dimensional spaces. But you'll have to find the dimension of $V$. – Jair Taylor Nov 05 '19 at 01:42
  • “If they have the same dimension, then they are isomorphic” is not true. For example, the finite vector spaces $\mathbb Z_2^2$ and $\mathbb Z_3^2$ are not isomorphic even though they are both two-dimensional. The two spaces must be defined over the same field (or at least over isomorphic fields). – amd Nov 05 '19 at 02:05

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For instance, $\textsf V$, which is not equal to $\textsf{M}_{2\times2}(\Bbb R)$, has not dimension $4$. Since $\textsf V$ can be written as $$ \left\{ \begin{pmatrix} a&b\\c&d \end{pmatrix}\in\textsf{M}_{2\times2}(\Bbb R):\, a+d=0 \right\} = \left\{ \begin{pmatrix} -d&b\\c&d \end{pmatrix}:\, b,c,d\in\Bbb R \right\} $$ it follows that $\dim(\textsf V)=3$. So, $\textsf{V}$ and $\textsf{W}$ are not isomorphic since they are not have the same dimension. Also, I don't know why you are so worried about a better way to determine when two spaces are isomorphic. I see this way quite simple.

azif00
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