If $\lambda$ is an eigen value of the matrix A with corresponding Eigen vector X, prove that $\lambda^n$ is an Eigen value of $A^n$ with corresponding Eigen vector X.

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written 9.4 years ago by

teamques10 ★ 70k

modified 3.9 years ago by

phenjoisilab • 0

engineering mathematics

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written 9.4 years ago by

teamques10 ★ 70k

Since $\lambda$ is an eigen value of A if X is the corresponding eigen vector

$AX = \lambda X$

Pre-multiplying by A

$A AX = A \lambda X = \lambda AX \\ \therefore A^2 X = \lambda AX = \lambda (\lambda X) = \lambda^2 X $

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

Continuing in this way

$A^n X = \lambda^n X$

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

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