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

Expert Answers info

txmedteach eNotes educator | Certified Educator

calendarEducator since 2012

write340 answers

starTop subjects are Math, Science, and Business

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!

check Approved by eNotes Editorial

txmedteach eNotes educator | Certified Educator

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.)

Unlock This Answer Now


check Approved by eNotes Editorial