A special case where similar matrices preserves linearly independence. $\{W_1, W_2, W_3\}$ forms a basis for $V=\{P|P^T=-P\}$ where $W_1 = \begin{bmatrix} 0& 0 &0\\ 0& 0 &1\\ 0 &-1 &0\end{bmatrix}; W_2  = \begin{bmatrix} 0& 0 &1\\ 0& 0 &0\\-1 &0 &0\end{bmatrix}, W_3  = \begin{bmatrix} 0 &1 &0\\ -1& 0& 0\\0 &0 &0\end{bmatrix}$ 
$\{v_1,v_2,v_3\}$ is any orthonormal basis for $\mathbb R^3$ which is used to construct a $3\times 3$ orthogonal matrix $M=(v_1,v_2,v_3).$ The question is to show that $Y=\{MW_1 M^{-1},MW_2 M^{-1},MW_3 M^{-1}\}$ also forms a basis for vector subspace $V=\{P|P^T=-P\}.$
As I obtained, $MW_1 M^{-1}=-v_3v_2^T+v_2v_3^T$, $MW_2 M^{-1}=-v_3v_1^T+v_1v_3^T$, $MW_3 M^{-1}=-v_2v_1^T+v_1v_2^T$. My idea is to show 
1). $-v_3v_2^T+v_2v_3^T$, $-v_3v_1^T+v_1v_3^T$ and $-v_2v_1^T+v_1v_2^T$ are linearly independent. 
2). $Span(-v_3v_2^T+v_2v_3^T, -v_3v_1^T+v_1v_3^T, -v_2v_1^T+v_1v_2^T)=V$
My question:
1). Since $v_1,v_2,v_3$ are orthogonal to each other, can I directly draw the conclusion that $-v_3v_2^T+v_2v_3^T$, $-v_3v_1^T+v_1v_3^T$ and $-v_2v_1^T+v_1v_2^T$ are linearly independent?
2).How can I show that $Span(-v_3v_2^T+v_2v_3^T, -v_3v_1^T+v_1v_3^T, -v_2v_1^T+v_1v_2^T)=V$ under that case that $v_1,v_2,v_3$ are not specific?
 A: Regarding the two questions you specifically ask: I see no way to directly draw the conclusion for 1.  The only proof I can think of for 1 or 2 amounts to a reformulation of the proof I outline below.
One approach to the original question (along the lines of Dietrich's comment) is as follows: show that the map $\phi: X \mapsto MXM^T$ defines a linear map from $V$ to $V$, and that this map is invertible (since $M$ is orthogonal, $M^{-1} = M^T$).  It immediately follows that if $W_1,W_2,W_3$ form a basis of $V$, then $\phi(W_1), \phi(W_2), \phi(W_3)$ form a basis of $V$.
An easy way to show that $\phi$ is invertible is to note that the map $\psi: X \mapsto M^TXM$ satisfies $(\psi \circ \phi)(X) = X$.

To get an answer to your second question with my approach, we can note the following.  If $M = (v_1,v_2,v_3)$ is invertible (but not necessarily orthogonal), then the elements whose span you consider are $\phi(W_1),\phi(W_2),\phi(W_3)$, with $\phi$ as defined above.
This time, the inverse to the map $\phi$ has the formula $\psi:X \mapsto M^{-1}XM^{-T}$.

Regarding the fact that $\phi$ is invertible implies that $\phi(W_1), \phi(W_2), \phi(W_3)$ form a basis of $V$:
To see that they are linearly independent, note that
$$
a_1 \phi(W_1) + a_2 \phi(W_2) + a_3 \phi(W_3) = 0 \implies\\
\phi(a_1 W_1 + a_2 W_2 + a_3 W_3) = 0 \implies\\
a_1 W_1 + a_2 W_2 + a_3 W_3 = 0 \implies\\
a_1 = a_2 = a_3 = 0.
$$
To see that they span $V$: take any $X \in V$.  We can select $a_1,a_2,a_3$ so that $\phi^{-1}(X) = a_1 W_1 + a_2 W_2 + a_3 W_3$. It follows that
$$
X = \phi(\phi^{-1}(X)) = 
\phi(a_1 W_1 + a_2 W_2 + a_3 W_3) = 
a_1 \phi(W_2) + a_2 \phi(W_2) + a_3 \phi(W_3).
$$
