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.

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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 $(X_AS)$.

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Wouldn't it be $X_A S$ instead of $SX_A$? –  Michael Hardy Apr 22 '12 at 0:05
And $(X_A)SBS^{-1}(X_A^{-1}) = (X_A S) B ( X_A S)^{-1}$. I'd have been explicit about that last step. –  Michael Hardy Apr 22 '12 at 0:07
Yeah sounds good. It would be $X_AS$ then. –  quantum Apr 22 '12 at 0:12
Thanks for the head up guys. –  user38268 Apr 22 '12 at 0:15

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

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