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I'm currently studying by the book "Theory of Lie groups", C. Chevalley. On page 6, paragraph before Proposition 6, he says: "The sets $M^S$, $M^{sh}$, $M^R$, $M^S$$\cap$$M^{sh}$, $M^R$$\cap$$M^S$, $M^R$$\cap$$M^{sh}$, $M^R$$\cap$$M^S$$\cap$$M^{sh}$, $M^s$ may all be considered as vector spaces over the field $R$ of real numbers; as such, their dimensions are $2n^2 - 2$, $n^2$, $n^2$, $n^2 - 1$, $n^2 - 1$, $n^2 - 1$, $n(n-1)/2$, $n(n-1)/2$ and $n(n-1)$ respectively."

$M^S$ is the set of matrices of trace O. $M^{sh}$ is the set of skew-hermitian matrices. $M^R$ is the set of real matrices. $M^s$ is the set of symmetric matrices.

I didn't understand how he concluded that each subset has the respective dimension he says.

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    I don't, either. Surely the $2\times2$ symmetric matrices form a vector space of dimension 3, not 2. But maybe there's some context that has been omitted. – 2012-08-18
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    The word "respectively" implies the two lists should have the same length, yet the first one has $8$ expressions and the second one $9$. But @GerryMyerson still has a point that neither the eight expression $\binom n2$ of the second list nor the final expression $n(n-1)$ seems to give the proper dimension of a space of symmetric matrices, whether real-symmetric ($n(n+1)/2$), complex-symmetric ($n(n+1)$) or complex self-adjoint ($n^2$) was meant. – 2012-08-18
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    Sorry, I've put three "n²-1". Actually, there are just two. I didn't understand it too, but this is how it is written =/ – 2012-08-19

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