Is a tensor which is symmetric in two indices still symmetric after raising/lowering one index? I have had this question for a while. I have yet to find information on this online or use this property in any calculation. I believe myself to have proven that it will still be symmetric but I am somewhat unsure of my proof. 
Suppose you have $A^{ij}=A^{ji}$, then,
$$ A^{ij}=A^{i}_{\phantom{i}k}g^{kj}=A^{\phantom{k}i}_{k}g^{kj}=A^{ji}\tag{1}$$
So $$A^{i}_{\phantom{i}k}g^{kj}=A^{\phantom{k}i}_{k}g^{kj}\tag{2}$$
Which implies $$A^{i}_{\phantom{i}k}=A^{\phantom{k}i}_{k}.\tag{3}$$
Is this true?
 A: It is symmetric but you actually have not proven it. In the first line you are just showing your initial hypothesis and in the second line you use what you are supposed to prove. The proof is simply
$$A^i_{\ \ j}=g_{jk}A^{ik}=g_{jk}A^{ki}=A_j^{\ \ i}.$$
A: TL;DR: No, it is per definition not symmetric, although OP's eq. (3) indeed holds. 
In more detail, if 
$$S~:=~S^{ij}~e_i \otimes e_j~\in~ T^2V~=~V\otimes V \tag{A}$$ 
and 
$$g~:=~g_{ij}~e^{\ast i} \otimes e^{\ast j}~\in~ T^2V^{\ast}~=~V^{\ast}\otimes V^{\ast}\tag{B}$$ 
are symmetric tensors 
$$S^{ij}~=~S^{ji}\qquad\text{and}\qquad g_{ij}~=~g_{ji},\tag{C}$$
then the mixed tensor
$$ M~:=~ M^i{}_j~e_i \otimes e^{\ast j}~\in~ V\otimes V^{\ast},\tag{D}$$ 
given by 
$$ M^i{}_j~:=~S^{ik}g_{kj},\tag{E} $$
is not symmetric. For starters, the transposed tensor
$$ M^T~:=~ (M^T)_i{}^j ~e^{\ast i}  \otimes e_j 
~:=~M^j{}_i ~e^{\ast i}  \otimes e_j~\in~ V^{\ast}\otimes V\tag{F}$$
lives in a different space, so the potential symmetry condition 
$$M~=~M^T  \qquad\qquad(\longleftarrow \text{Wrong!}) \tag{G}$$ 
is formally meaningless. What holds instead is
$$ M^j{}_i~=:~(M^T)_i{}^j~=~g_{in} ~M^n{}_m~ (g^{-1})^{mj}~=:~M_i{}^j. \tag{H}$$
Example: If 
$$S^{ij}~=~\begin{pmatrix}0&1\cr1&0\end{pmatrix}
\qquad\text{and}\qquad g_{ij}~=~\begin{pmatrix}1&0\cr0&2\end{pmatrix},\tag{I}$$
then the matrix
$$ M^i{}_j~=~\begin{pmatrix}0&2\cr1&0\end{pmatrix}\tag{J}$$
is not symmetric.
