How to show that if a matrix A is diagonalizable, then a similar matrix B is also diagonalizable? So a matrix $B$ is similar to $A$ if for some invertible $S$, $B=S^{-1}AS$. My idea was to start with saying that if $A$ is diagonalizable, that means $A={X_A}^{-1}\Lambda_A X_A$, where $X$ is the eigenvector matrix of $A$, and $\Lambda$ is the eigenvalue matrix of $A$.
And I basically want to show that $B={X_B}^{-1}\Lambda_B X_B$. This would mean $B$ is diagonalizable right?
I am given that similar matrices have the same eigenvalues, and if $x$ is an eigenvector of $B$, then $Sx$ is an eigenvector of $A$. That is, $Bx=\lambda x \implies A(Sx)=\lambda(Sx)$.
Can someone enlighten me please? Much appreciated.

Suppose $A$ diagonalisable, $B$ similar to $A$.
Then we have that there is an invertible matrix $S$ such that $S^{-1}AS = B$. It follows that $SBS^{-1} =A$. Since $X_A$ is the eigenvector matrix for $A$ (if  that's what you call it), it follows that
$$\begin{eqnarray*} \Lambda_A &=& X_A A X_A^{-1} \\
&=& (X_A)SBS^{-1}(X_A^{-1})\\
&=& (X_AS)B(X_AS)^{-1} 
\end{eqnarray*}$$
is a diagonal matrix. Hence $B$ is diagonalisable, diagonalised by the matrix $(X_AS)$.
