Suppose that for the complex square matrices it holds
Prove that .
If , then is invertible. So we get
But this equality is impossible: taking trace on both sides, we get
We get a contradiction, which shows that is singular, i.e. .
Let be a natural number such that . Evaluate the power
This is a very standard exercise in diagonalisation of matrices and there would be no reason to post it here , if it did not include the Fibonacci result. We are proving that
where denotes the – th Fibonacci number. The proof now follows with an induction on .
Let be a Jordan measurable set of zero measure. Prove that .
Let be irreducible with degree . If has a root on the unit circle then is even and
Let be positive real numbers. Prove that
We apply the AM – GM inequality, thus:
Hence it suffices to prove that which holds because it is equivalent to .