With reference to I.N. Herstein Topics in Algebra theorem 6.3.2, it states that if $V$ a n-dimensional vector space over field $F$, and the linear transformation $T$ has the matrix $m_1(T)$ has the basis $v_1, ..., v_n$ and the matrix $m_2(T)$ in the basis of $w_1,...,w_n$, then there is a matrix C such that $m_2(T)=Cm_1(T)C^{-1}$. This is oddly similar to the diagonalisation of a matrix where $A=PDP^{-1}$. In this case, I know that the matrix $D$ is the matrix of eigenvalues of $A$, which is in the basis of $v_1, v_2, v_3$, the corresponding eigenvectors of $A$, and this is identical to the definition of $m_1(T)$ in the theorem.
I would like to show that $P$, the matrix of eigenvectors, satisfy the matrix $C$ in the theorem, that maps $v_1, ..., v_n$ to $w_1,...,w_n$ which I am not sure how, since I do not know which basis $A$ is in.