Definition of symmetric tensor is well defined? I know that what a symmetric tensor is. I saw an exercise that made me a bit dubious about symmetric tensor.
Suppose $S$ is a symmetric $(1,1)$ tensor. then $g(SX,Y)=g(X,SY)$. In index notation $S_i{\ }^j=S_j{\ }^i$.
But I saw somewhere that claimed

If a symmetric $(1,1)$ tensor is symmetric in its indices with respect to  every basis then it is a multiple of identity tensor. i.e. $S_i{}^j=\alpha\delta_i{}^j$.

Isn't the indices with respect to  every basis symmetric always? ($S_i{\ }^j=S_j{\ }^i$) If the answer is no  then I think the definition of symmetric  tensor is not well defined. Is it?
 A: You can actually see what happens already on the level of linear algebra. For a symmetric bilinear form $b:V\times V\to\mathbb R$ there is an obvious notion of being symmetric. This corresponds to the fact that for each basis $\{v_i\}$ for $V$, the matrix $(b(v_i,v_j))_{i,j=1,\dots,n}$ is symmetric. Correspondingly, there is a natural notion of symmetry for $\binom02$-tensor fields on manifolds and this indeed amounts to $S_{ij}=S_{ji}$ with respect to any local frame.
In the presence of an inner product, you can convert a linear map $f:V\to V$ to a symmetric bilinear form $b:V\times V\to\mathbb R$ via $b(v,w)=\langle f(v),w\rangle$. Then symmetry of $b$ is equivalent to $\langle f(v),w\rangle=\langle v,f(w)\rangle$. However, the matrix $b(v_i,v_j)$ coincides with the matrix representation of $f$ with respect to the basis $\{v_i\}$ only if the basis orthonormal. Otherwise, you would have to use the dual basis with respect to the inner product in one entry of $b$. So symmetry of the linear map is equivalent to symmetry of a matrix representation with respect to an orthonormal basis but not with respect to a general basis. There is a counterpart to this on Riemannian manifolds, and the "good" condition in the Riemannian setting is that $g_{ik}S^k_j$ is symmetric in $i$ and $j$.
A: There is some subtlety about this question which I'll try to address. Let $V$ be an $n$-dimensional real vector space and let $S$ be a "$(1,1)$-tensor". Since there are various different objects which can be identified with $(1,1)$-tensors (linear maps on $V$, linear maps on $V^{*}$, bilinear maps $V \times V^{*} \rightarrow \mathbb{R}$, and so on) let's be precise about what we mean and take $S$ to be a bilinear map $S \colon V \times V^{*} \rightarrow \mathbb{R}$. Then, given a basis $(e_i)$ for $V$ (with $(e^j)$ the corresponding dual basis), we can construct
the square matrix $S_i{\ }^j := S(e_i, e^j)$. Note that the placement of indices tells us what type of object $S$ is.
Now, we can try and define $S$ to be "symmetric" if we have $S_i{\ }^j  = S_j{\ }^i $ for all $1 \leq i, j \leq n$. However, this turns out to be ill-defined as it is possible for this condition to hold with respect to one basis but not the other. This is not surprising: If you think about the tensor $S$ itself (and not the representing matrix), you see that it takes two arguments of different type (one in $V$, one in $V^{*}$) so a "symmetry condition" such as $S \left( v, \varphi \right) = S \left( \varphi, v \right)$ (where $v \in V$ and $\varphi \in V^{*}$) doesn't even compile. This is also reflected in Einstein's notation where in the equation
$S_i{\ }^j  = S_j{\ }^i$ one compares components where $i$ is both an upper and a lower index which represent different things.
Now let's assume that $V$ is endowed with an inner product $\left< \cdot, \cdot \right>$ which is in particular a $(0,2)$-tensor represented by $g_{ij}$ with respect to some basis. Then you can use the metric to lower the $j$ index in $S_i{\ }^j$ to obtain $S_{ij} := S_i{\ }^k g_{kj}$. The resulting array $S_{ij}$ with two lower indices represents the $(0,2)$-tensor $\hat{S} \colon V \times V \rightarrow \mathbb{R}$ given by $\hat{S} \left( v, w \right) = S \left( v, u \mapsto \left< w, u \right> \right)$. Then you can say that $S$ is symmetric if $S_{ij} = S_{ji}$ for all $1 \leq i, j \leq n$. This corresponds to the fact that $\hat{S}$ is symmetric in the sense that $\hat{S} \left( v, w \right) = \hat{S} \left( w, v \right)$ which now makes sense since both inputs of $\hat{S}$ are of the same type. Explicitly, you say that $S$ is symmetric if
$$ S_{ij} := S_i{\ }^k g_{kj} = S_j{\ }^k g_{ki} := S_{ji}$$ for all $1 \leq i,j \leq n$ and the representing arrays are taken with respect to the same basis $e_i$. Now, the important thing is that if this condition holds for one basis, then it holds for all bases, not only orthonormal ones! However, only when the basis is orthonormal we have $g_{ij} = \delta_{ij}$ and so $S_{ij} = S_i{\ }^j$ and $S_{ji} = S_j{\ }^i$ and hence you can deduce that if $S$ is symmetric then it is represented by a symmetric matrix in an orthonormal basis.
So let me summarize. Let $S$ be a $(1,1)$-tensor so it can be represented naturally only by a "mixed-upper-lower-indexed-array" $S_i{\ }^j$. If there is no metric, one has no notion of symmetry of $S$. If there is a metric around, we can convert between the two types of indices. In which case, there is a notion of symmetry and we have:

*

*$S$ is symmetric iff $\hat{S}$ is symmetric iff $S_{ij} = S_{ji}$ with respect to some basis iff $S_{ij} = S_{ji}$ with respect to every basis iff
$S_i{\ }^j = S^j_{{\,\, }i}$ with respect to every basis (I'll let you figure out the meaning of the last equality by yourself).

*$S$ is symmetric iff $S_i{\ }^j  = S_j{\ }^i$ with respect to some orthonormal basis iff $S_i{\ }^j  = S_j{\ }^i$ with respect to every orthonormal basis.

