Represent a vector as linear combination of eigen-vectors: $$u_0=c_1x_1+....c_nx_n$$ Now, $$Au_0=c_1\lambda_1x_1+....c_n\lambda_nx_n$$ where $\lambda$ is eigen-value. $Ax_i=\lambda_ix_i$ since $\lambda_i$ and $x_i$ are eigen-value and vector.

Why is this true? $$A^{100}u_0=c_1\lambda_1^{100}x_1+....c_n\lambda_n^{100}x_n$$

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    $\begingroup$ What does $A^2 u_0=A(Au_0)$ look like? $A^3 u_0$? Once you know the pattern, you can prove by induction. $\endgroup$ – Semiclassical Jul 14 '16 at 14:13

You have: $$Au_0=c_1\lambda_1x_1+....c_n\lambda_nx_n,$$

then $$A^2u_0=A(Au_0)=c_1\lambda_1A(x_1)+....c_n\lambda_nA(x_n)=c_1\lambda_1^2x_1+....c_n\lambda_n^2x_n.$$

Now repeat the same argument other $98$ times.


The way you're working the easiest way to go is by induction:

$$A^2u_0=A(c_1\lambda_1x_1+...+c_n\lambda_nx_n)=\sum_{k=1}^nc_k\lambda_k Ax_k=\sum_{k=1}^nc_k\lambda_k^2x_k$$

and etc.

Using similarity we could show the above with diagonal matrices and stuff.


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