Question

show that the sum of the roots of the charactristic c equation of a square matrix A is the trace of A​

Answer 1

Answer:

Step 1. If A and B are two n×n matrices then Tr(AB)=Tr(BA).

Step 2. If A is an n×n complex matrix, then there is a triangular matrix T, and an invertible matrix P such that A=P−1TP, hence

Tr(A)=Tr(P−1TP)=Tr(PP−1T)=Tr(T)

But the diagonal elements of T are the eigenvalues of A.

Remark. This works on any field if the characteristic polynomial can be split into the product of first degree polynomials. And the result is wrong otherwise.

Answer 2

It is direct if you use Schur's decomposition

That is, for any square matrix AA with complex entries, there exist unitary matrix UU such that U−1A UU−1A U is an upper triangular matrix with eigenvalues in the diagonal.