# Proof the Similarity of Matrices

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 $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix}$$

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.
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 $$\lambda^3-6\lambda^2+11\lambda-6=0=(\lambda-1)(\lambda-2)(\lambda-3).$$