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.
Assuming the usual Lie algebra operations on $M_n(\Bbb R)$, if $A$ and $B$ are symmetric:
$A^T = A, \tag 1$
$B^T = B, \tag 2$
$[A, B]^T = (AB - BA)^T = (AB)^T - (BA)^T$ $= B^TA^T - A^TB^T = BA - AB = [B, A] = -[A, B], \tag 3$
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.