Hero 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. one year ago one year ago

1. malevolence19

$\left[\begin{matrix}\psi & \xi \\ \chi & \zeta\end{matrix}\right] \implies (\psi - \lambda)(\zeta - \lambda) - \xi * \chi = 0$ 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

Brilliantly done!

3. eliassaab

No, they are not the same always.

4. Hero

Hmmm

5. Hero

I need a counterexample.

6. Hero

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

7. eliassaab

\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

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

9. eliassaab

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

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

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

12. Zarkon

same rank

13. eliassaab

No, you cannot conclude that they have the same rank

14. eliassaab

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

15. Hero

So what exactly am I supposed to conclude?

16. Zarkon

that they have the same eigenvalues ;)

17. eliassaab

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

18. Hero

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

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

20. eliassaab

21. Zarkon

the trace will be the same

22. eliassaab

The trace is the sum of eigenvalues.

23. Zarkon

the trace is the sum of the main diagonal

24. eliassaab

but it is equal to the sum of the eigenvalues.

25. Zarkon

by theorem..not my definition

26. eliassaab

Of course.

27. Zarkon

that is what I was pointing out

28. eliassaab

I see/

29. eliassaab

@hero are you satisfied with what is posted above.

30. Hero

yes,thanks

31. eliassaab

yw

32. Hero

I can only award one medal

33. Zarkon

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

34. Hero

I agree

35. TuringTest

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

36. Hero

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

37. malevolence19

I love this conversation haha @TuringTest @Zarkon @eliassaab