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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

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  • $\begingroup$ How does a $3 \times 3$ matrix have "at least two degenerate eigenvalues"? To be degenerate, they must have multiplicity greater than one. Having at least two degenerate eigenvalues means, counted with multiplicity, there are at least four eigenvalues. A $3 \times 3$ matrix can have, counted with multiplicity, at most three eigenvalues. The eigenvalues are solutions of the (cubic) characteristic equation. $\endgroup$ – Fly by Night Apr 6 '15 at 21:51
  • $\begingroup$ Sorry, I used the word triple degenerate and double degenerate. I am not sure what the appropriate wording should be. $\endgroup$ – linuxfreebird Apr 6 '15 at 21:54
  • $\begingroup$ you want to use the word (algebraic) multiplicity, so some eigenvalue has multiplicity at least 2. $\endgroup$ – rVitale Apr 6 '15 at 22:17
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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 $(a_1,b_1),(a_2,b_2),(a_3,b_3)$

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  • $\begingroup$ Wow. I did not even consider the 3 possible roots. More specifically, my $M$ is a symmetric positive definite matrix. $\endgroup$ – linuxfreebird Apr 6 '15 at 21:49
  • $\begingroup$ Should I post a variation of this question, more specific to a positive definite symmetric matrix? $\endgroup$ – linuxfreebird Apr 6 '15 at 22:21
  • $\begingroup$ In that case (positive definite symmetric), all the eigenvalues need to be real and positive. You can still have multiple real and positive solutions to the system of equations above: for example when $t=4,d=1$ the cubic has two positive real solutions $b_1,b_2$ both yielding a positive real solution for $a$, call them $a_1,a_2$. You can check that $a_1 \neq b_1$,$a_2 \neq b_2$. $\endgroup$ – rVitale Apr 6 '15 at 22:40
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If $\text{tr}(M)=-6$ and $\det(M)=-8$, then you could have eigenvalues $1,1,-8$ or $-2, -2, -2$

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  • $\begingroup$ Wow. I did not even consider the 3 possible roots. More specifically, my $M$ is a symmetric positive definite matrix. $\endgroup$ – linuxfreebird Apr 6 '15 at 21:49

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