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Let ( Ain M_n(mathbb{R})hspace{2mm} ) and ( hspace{2mm}lambda_1, lambda_2,...,lambda_n hspace{2mm} )(no need distinct) be eigenvalues...

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Advance Math

Let ( Ain M_n(mathbb{R})hspace{2mm} ) and ( hspace{2mm}lambda_1, lambda_2,...,lambda_n hspace{2mm} )(no need distinct) be eigenvalues of A. Show that 

a). ( sum _{i=1}^nlambda _i=trleft(Aight) )  b). ( :prod _{i=1}^nlambda _i=left|Aight|: )

 

Answer & Explanation Solved by verified expert
4.4 Ratings (615 Votes)

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| )
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