Why the two following maps agree on the basis vector for $R^3$? I cannot manage to understand the proof of the theorem below, could you elaborate why the maps agree on the basis vectors in $R^3$? For $v = (v_x,v_y,v_z)^t \in \mathbb{R}^3$ the matrix $\hat{v}$ is defined as
$$
\hat{v} = \left(
\begin{array}{lll}
0 & -v_z & v_y \\
v_z & 0 & -v_x \\
-v_y & v_x & 0
\end{array}
\right)
$$


Lema 5.4. (The hat operator). For a vector $T\in\mathbb R^3$ and a matrix $K\in\mathbb R^{3\times 3}$, if $\det(K)=+1$ and $T'=KT$, then $\widehat T = K^T \widehat{T'} K$.
Proof. Since both $K^T\widehat{(\cdot)}K$ and $\widehat{K^{-1}(\cdot)}$ are linear maps from $\mathbb R^3$ to $\mathbb R^{3\times 3}$ one may directly verify that these two linear maps agree on the basis vectors $[1,0,0]^T$, $[0,1,0]^T$ and $[0,0,1]^T$ (using the fact that $\det(K)=1$).

This is Lemma 5.4 from An Invitation to 3-D Vision (Yi Ma et al.), page 113.
 A: $\newcommand{\Reals}{\mathbf{R}}$Here's another argument, not using a basis. First note that if $v \in \Reals^{3}$, then $\hat{v}$ represents the (minus) cross product in the sense that for all $w$ in $\Reals^{3}$,
$$
\hat{v} w = -v \times w.
\tag{1}
$$
If $u$, $v$, and $w$ are arbitrary vectors in $\Reals^{3}$ and $K$ is an arbitrary $3 \times 3$ real matrix, then
$$
Ku \cdot (Kv \times Kw) = (\det K) u \cdot (v \times w)
\tag{2}
$$
because the triple product $u \cdot (v \times w) = \det[u\ v\ w]$ is the (signed) volume of the parallelipiped spanned by the ordered triple $u$, $v$, $w$, and $\det K$ is the multiplicative factor for volume under multiplication by $K$.
Finally, by the definition of the dot product,
$$
Ku \cdot (Kv \times Kw) = u^{T}K^{T} (Kv \times Kw).
\tag{3}
$$
If $T' = KT$, then for all $u$ and $w$ in $\Reals^{3}$,
\begin{align*}
  u^{T}(\hat{T}w)
  &= u \cdot (\hat{T}w) && \\
  &= -u \cdot (T \times w) && (1) \\
  &= -Ku \cdot (KT \times Kw) && (2),\ \det K = +1 \\
  &= -u^{T} K^{T} (T' \times Kw) && (3),\ T' = KT \\
  &= u^{T} K^{T} \widehat{T'} K w && (1).
\end{align*}
Since $u$ and $w$ were arbitrary, the matrices in the middle are equal, i.e.,
$$
\hat{T} = K^{T}\widehat{T'}K.
$$
A: Take $T=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and
$$K=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}.$$Then$$\widehat{T}=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix}.$$On the other hand, $T'=K.T=\begin{pmatrix}a_{11}\\a_{21}\\a_{31}\end{pmatrix}$. So$$\widehat{T'}=\begin{pmatrix}0&-a_{31}&a_{21}\\ a_{13}&0&-a_{11}\\-a_{12}&a_{11}&0\end{pmatrix}$$and therefore$$K^T\,\widehat{T'}\,K=\begin{pmatrix}0&0&0\\0&0&-\det K\\0&\det K&0\end{pmatrix}=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix}=\widehat{T}.$$The computations are similar for the other vectors of the canonical basis.
Note that in order to compute the matrix $K^T\,\widehat{T'}\,K$, you only have to compute three of its nine entries, since it is necessarily an anti-symmetric matrix.
