$T \cdot W=0$ for every skew symmetric tensor $T$ implies that $T$ is a symmetric tensor I'm trying to prove by myself the following easy fact. The context is continuum mechanics (in particular I'm using Gurtin's book):

Let $T$ a tensor and $W%$ a skew-symmetric tensor. If $T \cdot W =0 $ for every $ W \in \text{Skw}$, then $T$ is symmetric.


I can't obtain $T=T^t$ by using the definition, but I can't.
$T \cdot W = \text{tr}(T^t W)=\text{tr}(T^t (W^t) ^t) =- \text{tr}(T^t W^t) = -\text{tr}(W^t T^t)=-W \cdot T^t$
so I have $$T \cdot W + W \cdot T^t =0$$
and I don't know how to get the thesis $T=T^t$. How should I move?
 A: Every linear operator $T$ is uniquely expressible as a combination $T=T_\mathrm{sym}+T_\mathrm{skew}$ of a symmetric operator and a skew-symmetric operator. Indeed, we can apply the adjoint to both sides and solve the system
$$ \begin{cases}
T & =T_\mathrm{sym}+T_\mathrm{skew} \\ T^\dagger &=T_\mathrm{sym}-T_\mathrm{skew}
\end{cases} $$
to yield the formulas
$$ T_\mathrm{sym}=\frac{1}{2}(T+T^\dagger), \qquad T_\mathrm{skew}=\frac{1}{2}(T-T^\dagger). $$
Not only is the vector space $\mathrm{End}(V)$ a direct sum of the subspaces of symmetric and skew-symmetric operators, this is an orthogonal direct sum. We verify orthogonality using the definition of the Frobenius-Schmidt inner product of two linear operators an an inner product space: $\langle T,W\rangle=\mathrm{tr}(T^\dagger W)$.
Suppose $T$ is symmetric and $W$ is antisymmetric. Then we may use $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ and $\mathrm{tr}(X^\dagger)=\mathrm{tr}(X)$ (if we're talking about real inner product spaces) to reason that $\langle T,W\rangle=0$. (Can you do this?)
If $T$ were not symmetric, it would have nonzero skew-symmetric part. What would $T\cdot W$ be then if we were to set $W=T_\mathrm{skew}$? Can it be $0$?
