# 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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### 2 Answers

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)`

where `vec(0)` is the zero vector of length n and `vec(x)` is the eigenvector of `A^2` corresponding to eigenvalue `lambda^2`.

In much the same way as constants, the above matrix expression can be factored, as long as we replace `I` with `I^2` :

`A^2-lambda^2I^2=(A-lambdaI)(A+lambdaI)`

**Side Note:** I know I don't feel comfortable factoring matrices unless I check whether the multiplication holds up the way I expect it to (because matrix multiplication not often commutative), so let's just check to make sure:

`(A-lambdaI)(A+lambdaI)=A(A)+AlambdaI-lambdaIA-lambdaIlambdaI`

I know that *multiplication with constants and the identity matrix is commutative*, though, so I can do some rearranging:

`A^2 + AlambdaI - AlambdaI - lambda^2I^2=A^2 - lambda^2I`

Looks like we were ok in terms of factoring the first expression!

**</Side Note>**

Now, we'll continue with our factored expression:

`(A+lambdaI)(A-lambdaI)vec(x) = vec(0)`

If we have a nontrivial solution for the eigenvector `vec(x)` , we know that `vec(x)` is not the zero vector. Therefore, either `A+lambdaI` or `A-lambdaI` is an nxn matrix containing only zeros. We can then separate the above equation into two possible cases:

1) `(A+lambdaI)vec(x) = vec(0)`

2) `(A-lambdaI)vec(x)=vec(0)`

If we simplify these two cases we get to possible equations:

1) `Avec(x) = -lambdavec(x)`

2) `Avec(x) = lambdavec(x)`

And there you have it: given that `lambda^2` is an eigenvalue of `A^2` we can say that either `lambda` orĀ `-lambda` is an eigenvalue of `A`.

**Sources:**

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!