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.

Hint: Substitute $A = X_A^{-1} \Lambda X_A$ into $B = S^{-1} A S$ and use the formula $D^{-1}C^{-1} = (CD)^{-1}$.
