I was reading some notes and there was the following section:
Start with a given vector $\vec{u}_0$. We can create a sequence of vectors in which each new vector is $A$ times the previous vector: $$\vec{u}_{k+1} = A\vec{u}_k$$ $\vec{u}_{k+1} = A\vec{u}_k$ is a first order difference equation, and $\vec{u}_{k} = A^k\vec{u}_0$ is a solution to this system.
We get a more satisfying solution if we write $\vec{u}_0$ as a combination of eigenvectors of $A$: $$\vec{u}_0 = c_1\vec{x}_1 + c_2\vec{x}_2 + · · · + c_n\vec{x}_n = S\vec{c}$$ Then: $$A\vec{u}_0 = c_1\lambda_1\vec{x}_1 + c_2\lambda_2\vec{x}_2 + · · · + c_n\lambda_n\vec{x}_n$$ and: $$\vec{u}_k=A^k\vec{u}_0=c_1\lambda_1^k\vec{x}_1 + c_2\lambda_2^k\vec{x}_2 + · · · + c_n\lambda_n^k\vec{x}_n=\Lambda^kS\vec{c}$$ (Here $S$ represents the matrix $S=[\vec{x}_1\quad \vec{x}_2\quad ...\quad \vec{x}_n]$, the matrix with the eigenvectors of $A$ as columns, and $\Lambda$ is a diagonal matrix with its non-zero entries being the eigenvalues of $A$.)
They later apply this to get a closed form equation for the $k$-th Fibonacci number, and the formula they get is correct, however on the last line of the excerpt I don't see why they get $$\vec{u}_k=\Lambda^kS\vec{c} \qquad (1)$$ instead of $$\vec{u}_k=S\Lambda^k\vec{c} \qquad (2)$$ Specifically because I don't see how using $(1)$ will get you that same summation as in the last line in the excerpt, which $(1)$ aims to compact. Moreover, if you assume $A$ has $n$ independent eigenvectors then using $A^k=S\Lambda^k S^{-1}$ you get $$\vec{u}_k=A^k\vec{u}_0=A^kS\vec{c}=S\Lambda^k S^{-1}S\vec{c}=S\Lambda^k\vec{c}$$ which is $(2)$, which is what I believe the matrix notation of that long summation would be.
Consider a new coefficient vector $$\eqalign{ \vec{b} &= \Lambda^k\,\vec{c} \cr }$$ or in component form $$\eqalign{ b_j &= \lambda^k_j\,c_j \cr\cr }$$ The $u$-vectors can be expressed in terms of the $S$-matrix as $$\eqalign{ \vec{u_0} &= S\vec{c} \cr \vec{u_k} &= S\vec{b} = S\Lambda^k\vec{c} \cr }$$