The power method is an [[Iterative Method]] to find the [[Dominating Eigenvalue]] of a [[Matrix]]. Let $A$ be an $n\times n$ [[Square Matrix]] that is [[Eigendecomposition|Diagonalizable]] with [[Eigenvector|Eigenvectors]] $\set{\vec\lambda_{1},\vec\lambda_{1},\dots}$ forming a basis of $n$, with respect to [[Eigenvalue|Eigenvalues]] $\set{\lambda_{1},\lambda_{2},\dots}$, assuming the eigenvalues are ordered. $ \huge |\lambda_{1}| \geq |\lambda_{2}| \geq \cdots \geq |\lambda_{n}| $ We start with a non zero initial guess $\vec x^{(0)}$. $ \huge \vec x^{(k+1)} = \frac{1}{\norm{ A\vec x^{(k)} } } A \vec x^{(k)} $ Note this method applies for any [[Norm]]. $\huge \begin{align} \vec x^{(0)} &= c_{1} \vec \lambda_{1} + c_{2} \vec \lambda_{2} + \cdots \\ \text{{with }} & c_{1},\dots,c_{n} \in \R \\ \end{align} $ $ \huge \begin{align} A^{k} \vec x^{(0)} &= c_{1} A^{k}\vec \lambda_{1} + \cdots + c_{n} A^{k}\vec \lambda_{n} \\ &= c_{1} \lambda_{1} \vec \lambda_{1} + \cdots + c_{n} \lambda_{n} \vec \lambda_{n} \\ \text{if } c_{1} \neq 0 &\wedge |\lambda_{1}| > |\lambda_{2}| \\ A^{k} \vec x^{(0)} &= \lambda_{1}^{k} \pa{ c_{1} \vec \lambda_{1} + \frac{\lambda_{2}^{k}}{\lambda_{1}^{k}} c_{2} \vec \lambda_{2} + \cdots + \frac{\lambda_{n}^{k}}{\lambda_{1}^{k}} c_{n} \vec \lambda_{n} } \end{align} $ Because $\lambda_1$ is dominant, all coefficients of the form $\frac{\lambda_{n}^{k}}{\lambda_{1}^{k}}$ will approach $0$ as ${ n \to \infty }$. If a given [[Eigenvalue]] $\lambda$ has [[Multiplicity]] higher than one, then that the method works - but this method struggles if two [[Eigenvalue|Eigenvalues]] have the same [[Absolute Value]]. A common appearance of this is having two eigenvalues that are [[Complex Conjugate|complex conjugates]] of one another. This method will often fail to converge if $\lambda_{1} \in \C$ and $\lambda_{2}=\lambda_{1}^{*}$. >[!example] >An example of what this sequence will do to a matrix whose dominating eigenvalues are conjugates. >![[Pasted image 20260323155613.png|500]]