written 5.9 years ago by
teamques10
★ 65k
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• modified 2.5 years ago |
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Low
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written 5.9 years ago by
teamques10
★ 65k
|
Let $\lambda$ be an eigen value of $A^n$.
Then there exists $X\neq0$ such that
$(A-\lambda I)X=0$
$AX-\lambda X=0$
$AX= \lambda X$ (1)
Post multiplying by A we get,
$A^2X=\lambda A X$
$A^2X=\lambda \lambda X$ from (1)
$A^2X=\lambda^2 X$
$\lambda^2$ is eigen value of $A^2$ and eigen vector corresponding to $\lambda^2$ …
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written 2.5 years ago by
teamques10
★ 65k
|
Does that still apply if n is negative number?
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