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Yes.

Is there a way to prove that that sequence is always lesser than 3 without e?

**Bezoux**- Replies: 3

Hi, everyone.

I had an interesting problem in class today. Find

Now, it's fairly obvious that the answer is 0, the question is how to prove it.

I had an immediate suggestion, obviously if

,

then the answer is 0. So, we had to prove that.

By definition,

Since both and M are positive, I can raise them to the degree of n, which is when I get , which is obviously correct for n>= a certain n0.

Now, my teacher told me this was wrong and the class ended before he could elaborate on why, and the test is next class, which worries me a bit. I do a lot of my proofs this way.

What she said is that we need to prove , after which we can easily use the squeeze theorem to solve the problem. Unfortunately, I've had no success in proving that statement.

Can you point out where I was wrong in my solution, or help me prove the teacher's statement? I keep getting stuck at , but the teacher doesn't want us to use e (or infinite geometric series, which would make the proof fairly straightforward by using the binomial formula).

Thank you in advance, I know I'm asking a lot.

**Bezoux**- Replies: 2

Hey everyone.

I have a very strange question that I know is true, but seems to be bugging me.

I define so that it doesn't include 0.

This function is obviously surjective, as in for any y from the codomain there is an x in the domain such that f(x)=y.

Is there a good way to prove it?

There might be a silly arithmetic mistake in there, but I think that was the idea.

Essentially, my first idea was to assume the opposite, i.e. that there is a polynomial such that p(A)=B, for example (I'd do the p(B)=A one separately).

Then,

I found a very useful link on finding the nth degree of a matrix (I can't link it unfortunately, but it's the first result that comes up when you google "finding the nth power of a matrix").

Still, something makes me think there's a much more elegant solution that I'm not seeing.

**Bezoux**- Replies: 10

Hi everyone,

I've encountered a problem while studying matrices.

A={{2,1},{3,-1}}, B={{4,-2},{3,-1}}

Prove that there does not exist a polynomial with real coefficients such that p(A)=B or p(B)=A.

I've read up on eigenvalues, eigenvectors, characteristic polynomials and diagonalization, but nothing seems to be making sense, as the whole thing gets way too complicated for a high school problem.

I think there's a way to do this without using any of the aforementioned.

Can you help me out, please?

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