Let with . If for every matrix then prove that .
Suppose that , say for some . Let be any permutation matrix with and let be the matrix obtained from by changing its -entry to . Finally let where .
We have that and that is a polynomial in . Furthermore, the coefficient of of this polynomial is depending on the sign of the corresponding permutation. So the polynomial is not identically zero, a contradiction.
Let . Define the group
Prove that where is the dihedral group.
Using or equivalently we can write each element of in the form where . Using we may assume that . Using we may also assume that . It is easy to prove inductively that .
Let . We prove that . Obviously . Furthermore, since
If (such that ) then hence and therefore or . If then hence which is a contradiction since .
which is precisely the dihedral group with elements.
Consider the matrices and . If is invertible prove that is also invertible.
So we have to answer the question if is a zero of the essentially same characteristic polynomials. and have quite similar characteristic polynomials. In fact if denotes the polynomial of , then the polynomial of will be . It is easy to see that cannot be an eigenvalue of the matrix, otherwise it wouldn’t be invertible. Now, let us assume that is not invertible. Then it must have an eigenvalue of and let be the corresponding eigenvector. Hence:
meaning that has an eigenvalue of which is a contradiction. The result follows.
Let denote . Consider and let be its eigenvalues. Prove that:
Give an example of a real matrix that is not normal but is. Exponentiation is to be taken with respect to the usual definition.