The real linear space of matrices $$A=\begin{bmatrix}a&-c\\c&a\end{bmatrix},$$ with $a$ and $c$ real numbers, which satisfy the conditions (i) $\det(A)\geq0$ and (ii) $\det(A)=0\to A=0$, is a representation of the complex numbers. Is that true for all (special) spaces of $n\times n$ matrices satisfying these two conditions?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
6
No: consider the 3-dimensional space of matrices $A=\begin{pmatrix} 0 & a & b & c\\ -a & 0 & c & -b\\ -b & -c & 0 & a\\ -c & b & -a & 0 \end{pmatrix}$. Then $\det(A)=\operatorname{Pf}(A)^2=(a^2+b^2+c^2)^2 $.
-
1$\begingroup$ Or for that matter the Hamilton quaternions, which are obtained from your $A$'s by adding an arbitrary multiple of the identity. $\endgroup$ Commented Aug 31, 2017 at 17:41
-
$\begingroup$ @NoamD.Elkies Could one have octonions also, and that's all? $\endgroup$ Commented Aug 31, 2017 at 18:10
-
$\begingroup$ @efferrari You can take any linear subspace of the quaternions, or various collections of skew-symmetric matrices using the Pfaffian as in the answer. Octonions are not associative and thus not realizable as a matrix ring. $\endgroup$– KimballCommented Sep 1, 2017 at 2:09
-
1$\begingroup$ @efferrari There are many more examples. For instance, a vector space with a positive quadratic form, say of dimension $n$, acts by multiplication on its Clifford algebra; this gives a $n$-dimensional vector space of $2^n\times 2^n$ matrices with your property. $\endgroup$– abxCommented Sep 1, 2017 at 3:26
-
$\begingroup$ @Kimball Thank you! From the beautiful example given by abx, I think now that multiplication is not an issue, in the sense that multiplication by a complex number is equivalent to a dilatation and a rotation in the complex plane and in the given example the rotation is in the 3D space of vectors (a,b,c). The existence of such a symmetry is a crucial issue. $\endgroup$ Commented Sep 1, 2017 at 15:34