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Is it true for all matrices $A\in M_{4 \times 4}$ if they all have the eigenvalues $0, 1, 2, 3$ then these matrices are similar? I know that if matrices have the same eigenvalues, then they have the same characteristic polynomial, the same determinant and the same range. But how can we be sure that this is true for every matrix in $M_{4 \times 4}$ with these eigenvalues? What if there are two matrices that are exactly the same? Thanks.

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    $\begingroup$ If they are exactly the same they are trivially similar. $\endgroup$ – Sahiba Arora Jun 6 '17 at 19:48
  • $\begingroup$ If an $n$ by $n$ matrix has $n$ distinct eigenvalues then it is diagonalisable. Further, the entries in the diagonalised matrix are exactly the eigenvalues of the matrix. Hence what you stated is true. $\endgroup$ – Zestylemonzi Jun 6 '17 at 19:48
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Let $A,B\in \mathscr M_{4\times 4}$, matrices with eigenvalues $0,1,2,3$. Then they are both diagonalisable, since their eigenvalues are distinct. In particular, we can express $A$ and $B$ as $$ A=U^{-1}DU\quad\text{and}\quad B=V^{-1}DV, $$ where $D=\mathrm{diag}(0,1,2,3)$, and $U,V$ non-singular matrices.

This implies that $$ A=U^{-1}DU=U^{-1}VV^{-1}DVV^{-1}U=(V^{-1}U)^{-1}B(V^{-1}U), $$ and hence $A$ and $B$ are similar.

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Recall the following:

Proposition. Two diagonalizable matrices are similar if and only if they have the same characteristic polynomial.

Proof. As you stated, two similar matrices have the same characteristic polynomial, since: $$\det(XI_n-PAP^{-1})=\det(P(XI_n-A)P^{-1})=\det(XI_n-A).$$ Conversely, let $A$ and $B$ be two diagonalizable matrices with the same characteristic polynomial. There exists $P_A,P_B\in\textrm{GL}_n$ and $D_A:=\textrm{diag}(\lambda_1,\ldots,\lambda_n),D_B:=\textrm{diag}(\mu_1,\ldots,\mu_n)$ such that: $$A=P_AD_A{P_A}^{-1}\textrm{ and }B=P_BD_B{P_B}^{-1}.$$ Hence, $D_A$ and $D_B$ have the same characteristic polynomial. Therefore, there exists $\sigma\in\mathfrak{S}_n$ such that: $$\lambda_i=\mu_{\sigma(i)}.$$ In particular, $D_A$ and $D_B$ are similar. Whence the result. $\Box$

Whence the result, since $4\times 4$ matrices that have $4$ distinct eigenvalues are diagonalizable.

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