show/hide this revision's text 5 specify different eigenvalue

As per my comment: you can definitely decide whether $n$ is even or odd, since it is if and only if $n$ is even that $A^2 = -I$ has a solution in an $n \times n$ real matrix $A$.

Here is how you can detect even complex dimension. If we have conjugation, we can define a real number. If we have Hermitian conjugation, we can define a Hermitian matrix to be one $H$ such that $H^* = H$; any one is diagonalizable (with real eigenvalues, not that it matters). One can say that $H$ has distinct eigenvalues: if $Hv = \lambda v$ and $Hw = \lambda w$, then $v = \mu w$ for some $\mu$. Then $n$ is even if and only if there is a Hermitian matrix $H$ with distinct eigenvalues and a matrix $A$ such that for every eigenvector $v$ of $H$ we have an eigenvector $w$ w$, with different eigenvalue, such that $Av = w$ and $Aw = -v$. In particular, (This describes $A$ has as having the matrix $\bigl(\vcenter{\overset{\begin{smallmatrix} 0 & \;-I \end{smallmatrix}}{\begin{smallmatrix} I & \;\hphantom{-}0 \end{smallmatrix}}}\bigr)$, written in the eigenbasis of $H$, so $n$ is even.H$.)
show/hide this revision's text 4 Fix fix.

As per my comment: you can definitely decide whether $n$ is even or odd, since it is if and only if $n$ is even that $A^2 = -I$ has a solution in an $n \times n$ real matrix $A$.

Here is how you can detect even complex dimension. If we have conjugation, we can define a real number. If we have Hermitian conjugation, we can define a Hermitian matrix to be one $H$ such that $H^* = H$; any one is diagonalizable (with real eigenvalues, not that it matters). One can say that $H$ has distinct eigenvalues: if $Hv = \lambda v$ and $Hw = \lambda w$, then $v = \mu w$ for some $\mu$. Then $n$ is even if and only if there is a Hermitian matrix $H$ with distinct eigenvalues and a matrix $A$ commuting with $H$ with $A^2 = -I$, and if such that for every eigenvector $v$ and $w$ are two eigenvectors of $H$ with real ratio, then we have an eigenvector $Av$ w$ such that $Av = w$ and $Aw$ also have real ratio.

(It is okay to be this explicit since Aw = -v$. In particular, $A$ has the matrix $\bigl(\vcenter{\overset{\begin{smallmatrix} 0 & \;-I \end{smallmatrix}}{\begin{smallmatrix} I & \;0 ;\hphantom{-}0 \end{smallmatrix}}}\bigr)$, written in the eigenbasis of $H$, does the trick.)so $n$ is even.
show/hide this revision's text 3 Fix construction

Here is how you can detect even complex dimension. FirstIf we have conjugation, you we can define a real number, since you . If we have complex Hermitian conjugationof scalars. Of course, you we can define a Hermitian matrix to be one $i$ by H$ such that $i^2 H^* = -1$ (up to a sign, anyway). You can also define a complex basisH$; from this, you can define a any one is diagonalizable (with real basis, by which I mean a sequence $e_1, \dots, e_n, i e_1eigenvalues, \dots, i e_n$ for which the $e_j$ are real-linearly independentnot that it matters). Of course, you can't mention Then $n$ but you can write is even if and only if there is a formula saying when Hermitian matrix $H$ and a complex basis is matrix $A$ commuting with $H$ with $A^2 = -I$, and if $v$ and $w$ are two eigenvectors of this form$H$ with real ratio, and then $n$ falls out. Thus, you can define a Av$ and $Aw$ also have real ratio.

(It is okay to be this explicit since the matrix $\bigl(\vcenter{\overset{\begin{smallmatrix} 0 & \;-I \end{smallmatrix}}{\begin{smallmatrix} I & \;0 \end{smallmatrix}}}\bigr)$, written in this basis, and the previous paragraph applies.eigenbasis of $H$, does the trick.)
show/hide this revision's text 2 Cover complex case.
show/hide this revision's text 1