Question: If lambda is eigen value of A then prove that (lambda)^n is eigen value of A^n

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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 $\lambda^2$ is X.

Again pre multiplying by A,

$A^3X=\lambda^2 A X$

$A^3X=\lambda^2 \lambda X$

$A^3X=\lambda^3 X$

$\lambda^3$ is eigen value of $A^3$ and X is corresponding eigen vector. If we proceed in same manner then, we get,

$A^nX=\lambda^nX$

$\lambda^n$ is eigen value of $A^n$ and X is corresponding eigen vector.

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