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Assuming that they are both Hermetian, positive definite and have the same full rank. (To show the converse that if two matrices have the same eigenvalues, they must have the same determinant is easy.) However, I want to know if there is an good way to show that if they two matrices have the same determinant and same trace it does not imply that they will have the same (real) eigenvalues. I tried to think of an quick example but came up only with one where the determinant is zero. Any ideas?

Edit: I know it's true for $2 \times 2$ matrices which is easy to prove by the characteristics polynomial. However, it should not be for $3 \times 3$ matrices? What about a $3\times 3$ proof by contradiction? Note that the determinant should not be zero, as both matrices are positive definite by assumption, which excludes eigenvalues of zero. Also, we state nothing about the order of the eigenvalues.

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  • $\begingroup$ for $2\times 2$ matrices it is true as the only coefficients of the char polynomial is the trace and determinant $\endgroup$ – Dominic Michaelis Mar 24 '13 at 19:21
  • $\begingroup$ I thought of that. But then they are not full rank? I need positive definite non-singular matrices. $\endgroup$ – Majte Mar 24 '13 at 19:23
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    $\begingroup$ I guess what you're looking for then, is two multisets of $n$ complex numbers $\Lambda=\{\lambda_1,\ldots \lambda_n\}$, $\Delta=\{\mu_1,\ldots,\mu_n\}$ such that $\Lambda\neq\Delta$ but $0\neq\Pi_{i=0}^n\lambda_i=\Pi_{i=0}^n\mu_i$ and $\Sigma_{i=0}^n\lambda_i=\Sigma_{i=0}^n\mu_i$. $\endgroup$ – Dan Rust Mar 24 '13 at 19:23
  • $\begingroup$ @Majte I'm not saying those are counter examples. They don't even have the same trace. I just wanna know if you're counting multiplicties or not. $\endgroup$ – Git Gud Mar 24 '13 at 19:24
  • $\begingroup$ I think multiplicities are alright, as long they are not 0. $\endgroup$ – Majte Mar 24 '13 at 19:25
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Just try any random example. For instance, $A=\operatorname{diag}(1,3,5)$ and $B=\operatorname{diag}(4,\frac{5+\sqrt{10}}2,\frac{5-\sqrt{10}}2)$.

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  • $\begingroup$ well done :) how did you come up with this? $\endgroup$ – Majte Mar 24 '13 at 19:43
  • $\begingroup$ @Majte Pick the diagonal of $A$ at random. Then pick a diagonal entry of $B$ at random and see if there is a positive solution for the other two entries. $\endgroup$ – user1551 Mar 24 '13 at 19:47
  • $\begingroup$ Start with 3 random positive number and begin with the matrix with those as its diagonal. This gives you a trace and a determinant. Choose another random number for the $a_{11}$ position of your second matrix and then, using the two equations you get from the sum and product of the eigenvalues, this gives you a method of determining the last 2 numbers to place in the rest of the diagonal of the second matrix. $\endgroup$ – Dan Rust Mar 24 '13 at 19:47
  • $\begingroup$ ohh, well done.. I was thinking far to complex lol $\endgroup$ – Majte Mar 24 '13 at 19:50
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$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\;\;,\;\;\;\begin{pmatrix}0.5&0&1\\0&0.5&0\\\!\!\!-1&0&2\end{pmatrix}$$

If you define "positive definite" as a symmetric matrix s.t.... then the above doesn't work...

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  • $\begingroup$ they are not real.I should add Hermetian.. sorry, my fault, as I mistakenly defined positive definite as real. $\endgroup$ – Majte Mar 24 '13 at 19:38
  • $\begingroup$ What do you mean "they are not real"? Al their entries are real numbers! $\endgroup$ – DonAntonio Mar 24 '13 at 20:24
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Recall that the trace is the sum of the eigenvalues and the determinant is their product. To get positive definite require all the eigenvalues to be positive.

Now:

$x1+x2+x3=a$

$x1*x2*x3=b$

$x1,x2,x3>0$

Now just show that in general there are many solutions to this set of equations (or obtain a counter example by plugging in numbers, which is what user1551 did).

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