written 5.8 years ago by | • modified 2.2 years ago |
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Low
written 5.8 years ago by | • modified 2.2 years ago |
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Low
written 5.8 years ago by |
Let A be a symmetric matrix and also it is real.
$A=A\prime$ (1) and $A=\bar A$ (2)
Let $\lambda$ be any eigen value of A
To prove that $\lambda$ is real, i.e $\lambda= \bar \lambda$
As $\lambda$ is eigen value of A there exists $X\neq0$ such that
$(A-\lambda I) X=0$
$AX-\lambda X=0$
$AX=\lambda X$ (3)
Taking $\theta$ on both sides,
$(AX)^\theta=(\lambda X)^\theta$
$X^\theta A\theta=X^\theta \lambda^\theta$
$X^\theta(\bar A) \prime= (\bar \lambda)\prime X^\theta$
$X^\theta( A) \prime= (\bar \lambda) X^\theta$ from (2)
$X^\theta( A) = (\bar \lambda) X^\theta$ from (1)
Post multiplying by X on both sides,
$X^\theta( A) X = (\bar \lambda) X^\theta X $
$X^\theta( \lambda) X = (\bar \lambda) X^\theta X $ from (3)
$X^\theta( \lambda) X = (\bar \lambda) X^\theta X $
$(\lambda- \bar \lambda) X^\theta X=0$ as $X\neq0$ so, $X^\theta \neq0$ so $X^\theta X\neq0$
$\lambda-\bar \lambda=0$
$\lambda=\bar \lambda$
$\lambda$ is real
Eigen values of real symmetric matrix are real.