A subspace is just a vector space that is contained. You just gotta show that for both they are closed under addition and scalar multiplication. Also, that they contain the additive identity, but this is trivial for both of them.

1) This is simply the set of symmetric matrices.

Let A and B be symmetric matrices.

Addition:

and by definition.

A+B you just add compnantwise, so the ijth entry of A + B is (check)

Scalar multiplication.

A is a symmetric matrix so . Let k be a scalar from the field K.

so kA is a symmetric matrix.

(check)

Clearly the 0 matrix is symmetric every entry is 0.

(check)

1) is a subspace.

2)T is fixed and this is just the orbit of T under conjugation (This first statement is only true in the set of invertible matrices, but I never used invertibility in the proof, I meant to remove this statement actually, but forgot). You are looking at the set of all matrices A such that AT=TA call it W.

Let A and B be matrices in W. Then AT=TA and BT=TB

Addition

(A+B)T=AT +BT= TA + TB=T(A+B)

so A+B is in W

(Check)

Let k be an element in the field.

T(kA)=(Tk)A=k(TA)=k(AT)=(kA)T

so kA is in W.

(check)

T(0)=0=(0)T

so 0 is in W

It is a subspace.

Done and done.