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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
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

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malevolence19Best ResponseYou've already chosen the best response.0
\[\left[\begin{matrix}\psi & \xi \\ \chi & \zeta\end{matrix}\right] \implies (\psi  \lambda)(\zeta  \lambda)  \xi * \chi = 0\] 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.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
No, they are not the same always.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Well, by default they wouldn't be the same matrices
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
\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.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
The matrices that malevolence posted are not the same either are they?
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
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?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
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.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
For example they cannot be similar either. Can you construct a counter example?
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
No, you cannot conclude that they have the same rank
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
\[ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)\\ and \\ \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right) \] have the same eigenvalue but are not similar.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
So what exactly am I supposed to conclude?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
that they have the same eigenvalues ;)
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
The have the same determinant. They have the same characteristic polynomial. Something like that.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
the product of the eigenvalues is always equal to the determinalt for any nxn matrix
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
@Zakron answered you correctly.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
the trace will be the same
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
The trace is the sum of eigenvalues.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
the trace is the sum of the main diagonal
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
but it is equal to the sum of the eigenvalues.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
by theorem..not my definition
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
that is what I was pointing out
 one year ago

eliassaabBest ResponseYou've already chosen the best response.2
@hero are you satisfied with what is posted above.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
I can only award one medal
 one year ago

ZarkonBest ResponseYou've already chosen the best response.5
I think that if you are green (or purple) you should be able to give more than one medal.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I don't think I can elaborate much on what Zarkon said do you have a specific question?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
The question I had for you Turing had nothing to do with this question.
 one year ago

malevolence19Best ResponseYou've already chosen the best response.0
I love this conversation haha @TuringTest @Zarkon @eliassaab
 one year ago
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