A,B are diagonalizable matrix and their characteristic polynomials are the same.prove that $A$ and $B$ are similar

Question

let A,B are diagonalizable matrix in ${c^n}$ and their characteristic polynomials are the same. can we prove that $A$ and $B$ are similar?

Answer 1

$A$ and $B$ have same eigenvalues and hence same diagonal matrix $D$. Let $A=PDP^{-1}$ and $B=QDQ^{-1}$ where $P$ and $Q$ are invertible (just using change of basis concept). Thus we have $D=P^{-1}AP$. Substituting, $B=QP^{-1}APQ^{-1}=(PQ^{-1})^{-1}A(PQ^{-1})$ . Taking $PQ^{-1}=S$ we get $B=S^{-1}AS$ and hence similarity follows.

Answer 2

By assumption, $A$ is similar to $\left(\begin{array}{cccc}\alpha_1&0&\cdots&0\\0&\alpha_2&\cdots&0\\\vdots&\cdots&\cdots&\vdots\\0&\cdots&\cdots&\alpha_n\end{array}\right)$, and hence its characteristic polynomial is given by$$f_A(x)=\prod_{i=1}^n(x-\alpha_i).$$The matrix $B$ is similar to a diagonal matrix with $\beta_1,\ldots,\beta_n$ on the diagonal, and hence$$f_B(x)=\prod_{i=1}^n(x-\beta_i).$$Since the two polynomials are assumed to be the same, we have that $(\alpha_i)$ are the same as $(\beta_i)$, up to reordering, and the two matrices are indeed similar.