Jordan Canonical Form

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

a). \( \sum _{i=1}^n\lambda _i=tr\left(A\right) \) b). \( \:\prod _{i=1}^n\lambda _i=\left|A\right|\: \)

Solutions

Expert Solution

Solution

we have \( P(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}S_1\lambda^{n-1}++...+(-1)^{n-n}S_n\lambda^{n-n} \hspace{3mm} (1) \)

since \( S_1=tr(A) \hspace{2mm}and\hspace{2mm} S_n=|A| \)

\( +\hspace{2mm}\lambda_1+\lambda_2+...+\lambda_n\hspace{2mm} \) are eigenvalue of A ,then

\( \hspace{2mm}P(\lambda_i)=0\hspace{2mm},\forall i=1,2,3,...,n.\hspace{2mm}Then. \)

\( P(\lambda)=(-1)^n(\lambda-\lambda_1)(\lambda-\lambda_2)....(\lambda-\lambda_n) \hspace{2mm}(2) \)

from (1) and (2) :

Therefore

a). \( \sum _{i=1}^n\lambda _i=-\frac{b}{a}=-\frac{(-1)^{n-1}S_1}{(-1)^n}=S_1=tr(A) \)

b). \( \:\prod _{i=1}^n\lambda _i=\frac{(-1)^nS_n}{(-1)^n}=S_n=|A| \)

Answer

Therefore

a). \( \sum _{i=1}^n\lambda _i=-\frac{b}{a}=-\frac{(-1)^{n-1}S_1}{(-1)^n}=S_1=tr(A) \)

b). \( \:\prod _{i=1}^n\lambda _i=\frac{(-1)^nS_n}{(-1)^n}=S_n=|A| \)

SUN BUNRA answered 2 years ago

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