how come 1-i another eigan value?

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akash.dinkar12
asked
in Linear Algebra
May 11, 2019

1,163 views
6 votes

Best answer

Characteristic equation $|A-\lambda I|=0$

$\lambda_1=1+i \implies \lambda_2=1-i $

$\because$ all diagonal elements=0, $\Rightarrow \text{trace}=0$

Trace = sum of eigen values=0

$\implies\lambda_1+\lambda_2+\lambda_3=0$

$\implies (1+i)+(1-i)+\lambda_3=0$

$\implies \lambda_3=-2$

$|A|=\text{product of eigen values}=(1+i)(1-i)(-2)$

$\implies |A|=-4$

$\lambda_1=1+i \implies \lambda_2=1-i $

$\because$ all diagonal elements=0, $\Rightarrow \text{trace}=0$

Trace = sum of eigen values=0

$\implies\lambda_1+\lambda_2+\lambda_3=0$

$\implies (1+i)+(1-i)+\lambda_3=0$

$\implies \lambda_3=-2$

$|A|=\text{product of eigen values}=(1+i)(1-i)(-2)$

$\implies |A|=-4$