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Suppose $A$ is a $3\times 3$ matrix with entries in a field $F$ of characteristic $0$, and assume $\operatorname{Tr}A = 6$, $\operatorname{Tr}(A^2)=14$, and $\det A = 6$. ($\operatorname{Tr}$ denotes the trace.) Prove that $A$ is similar over $F$ to the diagonal matrix $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix} $$

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Let $x, y$ and $z$ denote the eigenvalues of $A$ over the algebraic closure of $F$. You are given that

$$x+y+z=6 \text{ and } x^2+y^2+z^2=14 \text{ and } xyz=6.$$ Solving this system for $x, y$ and $z$ you can find that the solutions are $1, 2$, and $3$ and since the eigenvalues are distinct the matrix can be diagonalized in the desired form.

A hint for solving the above system of equations:

By subtracting the square of the first equation from the second equation, you can see that $xy+yz+zx=11$. Thus the characteristic equation of $A$ is $$\lambda^3-6\lambda^2+11\lambda-6=0=(\lambda-1)(\lambda-2)(\lambda-3).$$

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