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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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    Thanks for the head up guys.2012-04-22
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Hint: Substitute $A = X_A^{-1} \Lambda X_A$ into $B = S^{-1} A S$ and use the formula $D^{-1}C^{-1} = (CD)^{-1}$.