Given matrix A with eigenvalue $\lambda$ and corresponding eigenvector x, prove $A^k$ has eigenvalue $\lambda^k$ for the same eigenvector x for any positive integer k.
Can I just use the eigenvalue definition Ax = $\lambda$x, divide x on both sides, then raising the power?
Best Answer
1.) $Ax = \lambda x$, given.
2.) Inductive hypothesis: $A^kx = \lambda^k x$ for some positive integer $k$.
3.) Operate on the equation of (2) with $A$: $A^{k + 1}x = A(A^k x) = A(\lambda^k)x = \lambda^k (Ax) = \lambda^{k + 1} x$.
4.) Conclude that $A^k x = \lambda^k x$ for all positive integers $k$.
QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!