If $T(v) \otimes w + v \otimes S(w)=0$ are the skew-symmetric operators $T$ and $S$ zero?

Question

Let $V,W$ be finite-dimensional real inner product spaces, and let $T \in \text{Hom}(V,V), S \in \text{Hom}(W,W)$ be skew-symmetric operators.

Define the following map $\phi \in \text{Hom}(V\otimes W,V\otimes W)$ by requiring $$ \phi(v \otimes w)=T(v) \otimes w + v \otimes S(w). $$ Now, suppose that $\phi=0$ is the zero element. Does $T=0,S=0$?

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If we remove the requirement of skew-symmetry, then this is not true, e.g. one can take $T=id, S=-id$.

I managed to prove this when $\dim V=\dim W=2$:

In that case every skew-symmetric map can be identified with $\begin{pmatrix} 0 & a \\\ -a & 0 \end{pmatrix}$, so the computation reduces to something easy.

Answer 1

Assume $\phi = T\otimes \text{id}_W + \text{id}_V\otimes S =0$.

Then, for $v,v' \in V, w \in W\backslash \{0\}$, we have $\langle w,Sw\rangle = 0$ and therefore \begin{align} 0 &= \langle v'\otimes w, \phi(v\otimes w)\rangle \\ &= \langle v'\otimes w, v\otimes Sw+ Tv\otimes w\rangle\\ &=\langle v',v\rangle \langle w,Sw\rangle + \langle v',Tv\rangle\|w\|^2 \\ &= \langle v',Tv\rangle\|w\|^2 \end{align}

Since this is true for all $v,v'\in V$, we have $T = 0$. By the same argument, $S = 0$.