this is a question from In V.I.Arnold's book about ordinary differential equations. the Q: Do the symmetric matrices form a Lie algebra with the same operation? What about the skew-symmetric matrices? .The question came after the definition:A vector space with a binary operation possessing properties 1, 2 , and 3 is called a Lie algebra. Thus vector fields with the operation of commutation form a Lie algebra. This operation is just as fundamental for all of mathematics as addition and multiplica tion.

  • $\begingroup$ What operation? $\endgroup$
    – Mark Sapir
    Sep 28 '20 at 4:41

Assuming the usual Lie algebra operations on $M_n(\Bbb R)$, if $A$ and $B$ are symmetric:

$A^T = A, \tag 1$

$B^T = B, \tag 2$

we have

$[A, B]^T = (AB - BA)^T = (AB)^T - (BA)^T$ $= B^TA^T - A^TB^T = BA - AB = [B, A] = -[A, B], \tag 3$

so $[A, B]$ is skew-symmetric and thus the symmetric matrices do not form a Lie algebra.

On the other had, with $A$ and $B$ skew,

$A^T = -A, \tag 4$

$B^T = -B, \tag 5$

$[A, B]^T = (AB - BA)^T = (AB)^T - (BA)^T$ $= B^TA^T - A^TB^T = (-B)(-A) - (-A)(-B)$ $= BA - AB = -[A, B], \tag 6$

which shows that $[A, B]$ is skew-symmetric, and thus that the skew-symmetric matrices do indeed form a Lie algebra.

The reader may easily extend this argument to showing that the skew-Hermitian matrices in $M_n(\Bbb C)$ form a Lie algebra, but that the Hermitian matrices do not.


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