## Hero Group Title If a 2x2 matrix has the same two eigenvalues as another 2x2 matrix, what conclusions can you infer about the two matrices? Make some hypotheses and attempt to prove them, or refute them via counterexamples. 2 years ago 2 years ago

1. malevolence19 Group Title

$\left[\begin{matrix}\psi & \xi \\ \chi & \zeta\end{matrix}\right] \implies (\psi - \lambda)(\zeta - \lambda) - \xi * \chi = 0$ Take the transpose of that matrix you have: $\left[\begin{matrix}\psi & \xi \\ \chi & \zeta\end{matrix}\right]^T=\left[\begin{matrix}\psi & \chi \\ \xi & \zeta\end{matrix}\right] \implies (\psi - \lambda)(\zeta - \lambda) - \xi * \chi = 0$ They are the same. I'm not sure if you can ALWAYS assume this but it seems pretty logical. I also don't know any theorems to name.

2. Hero Group Title

Brilliantly done!

3. eliassaab Group Title

No, they are not the same always.

4. Hero Group Title

Hmmm

5. Hero Group Title

I need a counterexample.

6. Hero Group Title

Well, by default they wouldn't be the same matrices

7. eliassaab Group Title

\left( \begin{array}{cc} 2 & 4 \\ 0 & 3 \\ \end{array} \right) \\ \left( \begin{array}{cc} 2 & 0 \\ 0 &3 \\ \end{array} \right) Have the same eigenvalues 2 and 3 but they are not the same.

8. Hero Group Title

The matrices that malevolence posted are not the same either are they?

9. eliassaab Group Title

Ok. I thought he wanted to conclude that if two matrices have the same eigenvalues then they are the same. This is not true in general. What do you think the answer to your question is?

10. Hero Group Title

Well, apparently, there's more than one conclusion. If you have two matrices that are not the same but have the same eigenvalues, I'm not really sure what to conclude which is why I posted the question.

11. eliassaab Group Title

For example they cannot be similar either. Can you construct a counter example?

12. Zarkon Group Title

same rank

13. eliassaab Group Title

No, you cannot conclude that they have the same rank

14. eliassaab Group Title

$\left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)\\ and \\ \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right)$ have the same eigenvalue but are not similar.

15. Hero Group Title

So what exactly am I supposed to conclude?

16. Zarkon Group Title

that they have the same eigenvalues ;)

17. eliassaab Group Title

The have the same determinant. They have the same characteristic polynomial. Something like that.

18. Hero Group Title

So any two matrices with the same eigenvalues will have the same determinant and characteristic polynomial? That doesn't seem accurate. Maybe for certain cases, but not for every case

19. Zarkon Group Title

the product of the eigenvalues is always equal to the determinalt for any nxn matrix

20. eliassaab Group Title

21. Zarkon Group Title

the trace will be the same

22. eliassaab Group Title

The trace is the sum of eigenvalues.

23. Zarkon Group Title

the trace is the sum of the main diagonal

24. eliassaab Group Title

but it is equal to the sum of the eigenvalues.

25. Zarkon Group Title

by theorem..not my definition

26. eliassaab Group Title

Of course.

27. Zarkon Group Title

that is what I was pointing out

28. eliassaab Group Title

I see/

29. eliassaab Group Title

@hero are you satisfied with what is posted above.

30. Hero Group Title

yes,thanks

31. eliassaab Group Title

yw

32. Hero Group Title

I can only award one medal

33. Zarkon Group Title

I think that if you are green (or purple) you should be able to give more than one medal.

34. Hero Group Title

I agree

35. TuringTest Group Title

I don't think I can elaborate much on what Zarkon said do you have a specific question?

36. Hero Group Title

The question I had for you Turing had nothing to do with this question.

37. malevolence19 Group Title

I love this conversation haha @TuringTest @Zarkon @eliassaab