Does the trace and determinant uniquely determine the eigenvalues of a 3 by 3 matrix with algebraic multiplicity of 2?

Question

I have a 3 by 3 matrix $M$ whose eigenvalues are $a$, $b$, and $b$.

The determinant and trace of $M$ are known from its eigenvalues: $det(M)=ab^2$ and $Tr(M)=2b+a$.

I wanted to show that if $Tr(M)=3 \ \sqrt[3]{det(M)}$, then $a=b$. I was told that this one equation is not sufficient to completely determine $M$'s eigenvalues to all be equal, if it is already know that it has at least algebraic multiplicity of 2.

Does the trace and determinant uniquely determine the eigenvalues of a 3 by 3 matrix with algebraic multiplicity of 2?

General Case

Answer 1

You have two variables and two equations, $d=ab^2$ and $t=a+2b$ where $d,t$ are some constants. Solve this system of equations:

$a=t-2b \rightarrow d=(t-2b)b^2 \rightarrow 2b^3-tb^2+d=0$

So you have a cubic in $b$ which has a generic solution with 3 roots $b_1,b_2,b_3$. Each of these determines a corresponding value for $a$, so in general your determinant and trace only determine the eigenvalues up to choosing a solution $(a_1,b_1),(a_2,b_2),(a_3,b_3)$

Answer 2

If $\text{tr}(M)=-6$ and $\det(M)=-8$, then you could have eigenvalues $1,1,-8$ or $-2, -2, -2$