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What is the order of the set of distinct (up to similarity) nxn matrices over $\mathbb{R}$ with determinant equal to some non-zero scalar... say 6? (eg. countable, uncountable etc.)

The set of matrices over $\mathbb{R}$ is uncountable. So is the order of the set consisting of classes of matrices with the same determinant. In each of those classes we have further subsets, the equivalence classes formed by grouping similar matrices. Each equivalence class of similar matrices represents one linear transformation expressed in terms of all possible bases of $\mathbb{R}^n$, so the size of each equivalence class is also uncountable.

What I'm not certian about is the size of the set of "different" transformations that have the same determinant. Is it uncountable too?

What can I put it in a correspondence with to show this?

Apologies if this is poorly worded-- let me know if there is a better way to ask this question.

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No apologies necessary. I think the question is very clearly worded. Minor nitpick: the similarity class of a scalar multiple of the identity matrix is not uncountable. –  Jonas Meyer Nov 22 '10 at 21:01

1 Answer 1

It has the same cardinality as $\mathbb{R}$ if $n\gt 1$. Consider diagonal matrices with entries $(x,6/x,1,1,\ldots,1)$, $x\neq0$. Two such are similar only in a situation where $\{x,6/x\}=\{y,6/y\}$, hence the conclusion on cardinality. (Of course $6$ is not special.)

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(The only additional detail worth mentioning is that ${}|{\mathbb R}|$ is an upper bound, and so this argument gives equality. Asaf's answer addresses this detail.) –  Andres Caicedo Nov 23 '10 at 2:34
    
@Andres: Agreed. I didn't know whether a little don would need this detail. –  Jonas Meyer Nov 23 '10 at 2:36
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Asaf's answer was deleted. Anyway, just in case: The set of $n\times n$ matrices with real entries is obviously in bijection with ${\mathbb R}^{n^2}$, which has the same size as ${\mathbb R}$. This gives the upper bound. –  Andres Caicedo Nov 23 '10 at 6:50

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