Solution
we have ( P(lambda)=(-1)^nlambda^n+(-1)^{n-1}S_1lambda^{n-1}++...+(-1)^{n-n}S_nlambda^{n-n} hspace{3mm} (1) )
since ( S_1=tr(A) hspace{2mm}andhspace{2mm} S_n=|A| )
( +hspace{2mm}lambda_1+lambda_2+...+lambda_nhspace{2mm} ) are eigenvalue of A ,then
( hspace{2mm}P(lambda_i)=0hspace{2mm},forall i=1,2,3,...,n.hspace{2mm}Then. )
( P(lambda)=(-1)^n(lambda-lambda_1)(lambda-lambda_2)....(lambda-lambda_n) hspace{2mm}(2) )
from (1) and (2) :
Therefore
a). ( sum _{i=1}^nlambda _i=-frac{b}{a}=-frac{(-1)^{n-1}S_1}{(-1)^n}=S_1=tr(A) )
b). ( :prod _{i=1}^nlambda _i=frac{(-1)^nS_n}{(-1)^n}=S_n=|A| )
Answer
Therefore
a). ( sum _{i=1}^nlambda _i=-frac{b}{a}=-frac{(-1)^{n-1}S_1}{(-1)^n}=S_1=tr(A) )
b). ( :prod _{i=1}^nlambda _i=frac{(-1)^nS_n}{(-1)^n}=S_n=|A| )
