Show the proof of this property of skew-symmetric matrix?

Question

For a vector $T \in \mathbb{R}^3$ and a matrix $K \in \mathbb{R}^{3x3}$, if $det(K) = +1$ and $T' = KT$, then $\hat{T} = K^T \hat{T'} K $, where the hat operator denotes the skew-symmetric matrix corresponding to the "hat-ed" vector.

Answer 1

In general, we have $\hat{T}v=T\times v$ for any $T,v\in\mathbb{R}^3$. Now let $v,w\in\mathbb{R}^3$ be arbitrary.

$<K^T \hat{T'}Kv,w>=<K^T (KT)\times Kv,w>=<KT\times Kv,Kw>$

By the scalar triple product formula, the right hand side is equal to $det(K[T,v,w])$, where $[T,V,w]$ is the matrix with $T$,$v$ and $w$ being its columns. But this is equal to $det(K)det([T,v,w])=det([T,v,w])=<T\times V,w>$. We have shown that $<K^T \hat{T'}Kv,w>=<T\times v,w>$ for any $w$, and since $w$ is arbitrary, we conclude

$$K^T \hat{T'}Kv=T\times v$$

But since $v$ was also arbitrary, we conclude $K^T \hat{T'}K=\hat{T}$.