# Do the symmetric matrices form a Lie algebra with the same operation? What about the skew-symmetric matrices?

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.

• What operation? 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.