How do we prove that the sum of the roots of the characteristic polynomial of a square matrix is equal to the trace of the matrix ? I want a proof which does not use much computation or determinants ; please help , Thanks in Advance .
[Math] To prove that the sum of the roots of the characteristic polynomial of a square matrix is equal to the trace of the matrix
eigenvalues-eigenvectorslinear algebramatrices
Best Answer
Step 1. If $A$ and $B$ are two $n\times n$ matrices then ${\rm Tr}(AB)= {\rm Tr}(BA)$.
Step 2. If $A$ is an $n\times n$ complex matrix, then there is a triangular matrix $T$, and an invertible matrix $P$ such that $A=P^{-1}TP$, hence $${\rm Tr}(A)={\rm Tr}(P^{-1}TP)={\rm Tr}(P P^{-1}T )={\rm Tr}(T)$$ 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.