A is nxn Matrix M(R) and lambda is a number in R such that lambda^2is an eigenvalue of A^2. Can we show that (+,-) lambda is eigenvalue of A? A is an element of Mn,n(R), with R = real numbers = field
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calendarEducator since 2012
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Wow, I'm sorry, I was reading through this, and I came up on a small mistake. Thankfully, it doesn't affect the proof.
`A+lambdaI` and `A-lambdaI` do not have to be matrices containing only zeros. However, that does not affect the problem, as you simply distribute the eigenvector `vec(x)` and subtract `+-lambdavecx` from both sides.
Again, I apologize, and I hope it didn't affect your understanding too much!
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calendarEducator since 2012
write340 answers
starTop subjects are Math, Science, and Business
In order for `lambda^2` to be an eigenvalue of `A^2` , we know that the following equation must hold:
`(A^2 - lambda^2I)vec(x) = vec(0)`
(The entire section contains 2 answers and 344 words.)
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