# Iterative Matrix Techniques Based on Averages
## 3 Computation of an Eigenvector of a Known Eigenvalue of a Linear Operator  
This section is devoted to studying the convenience of using the N-iteration to find an eigenvector/eigenfunction corresponding to a non-null known eigenvalue of a linear operator in finite or infinite dimensions. Let \(T:E\to E\) be a linear operator defined on a normed space \(E\). If \(\lambda\in\sigma(T)\), where \(\sigma(T)\) is the point spectrum of \(T\), the search for an eigenvector (or eigenfunction if \(E\) is a functional space) associated with \(\lambda\) consists in finding an element \(v\in E\) non-null such that \(Tv=\lambda v\). The solution of this equation can be reduced to a fixed-point problem considering that this equality is equivalent to  
\[\lambda^{-1}Tv=v,\]  
assuming that \(\lambda\) is non-null. The following result gives sufficient conditions for the convergence of the N-iteration applied to \(\lambda^{-1}T\) to find an eigenvector associated to a non-null eigenvalue. The next result was given in reference [22]. Proposition 1: _Let \(E\) be a normed space and \(C\subseteq E\) be nonempty, closed and convex. Let \(T:C\to C\) be a nonexpansive operator such that the set of fixed points is nonempty; then, the sequence of N-iterates \((f_{n})\) for \(f_{0}\in C\) is such that the sequence \((||f_{n}-f^{*}||)\) is convergent for any fixed point \(f^{*}\in C\) of \(T\). If \(E\) is a uniformly convex Banach space and the scalars are chosen such that \(0<\inf\alpha_{n}\leq\sup\alpha_{n}<1\) and \(0<\inf\gamma_{n}\leq\sup\gamma_{n}<1\), then the sequence \((||f_{n}-Tf_{n}||)\) tends to zero for any \(f_{0}\in C\)._  
Theorem 3.1: _Let \(E\) be a uniformly convex Banach space, and \(K:E\to E\) be a linear and compact operator. If \(\lambda\in\sigma(K)\) is non-null, such that \(||K||\leq|\lambda|\), then the N-iteration with the scalars described in the previous proposition applied to the map \(\lambda^{-1}K\) converges strongly to an eigenvector associated with \(\lambda\)._  
Proof: Since \(\lambda\in\sigma(K)\), there exists \(f^{*}\in E\), \(f^{*}\
eq 0\), such that \(Kf^{*}=\lambda f^{*}\). Then, \(f^{*}\) is a fixed point of \(\lambda^{-1}K\). The latter operator is nonexpansive since  
\[||\lambda^{-1}Kf-\lambda^{-1}Kg||\leq|\lambda^{-1}|||K|||f-g||\leq||f-g||,\]  
for any \(f,g\in E\), due to the hypotheses on \(\lambda\). Regarding invariance, if \(f\in\bar{B}(\bar{0},R)\) then  
\[||\lambda^{-1}Kf||\leq||f||\leq R,\]  
and \(\lambda^{-1}Kf\in\bar{B}(\bar{0},R)\). Let \(f_{0}\in\bar{B}(\bar{0},R)\), and \((f_{n})\subseteq\bar{B}(\bar{0},R)\) be the sequence of N-iterates of \(f_{0}\). Since \(\lambda^{-1}K\) is compact, there exists a convergent subsequence \((\lambda^{-1}Kf_{m_{k}})\). Let \(f\) be the limit of this subsequence. Then,  
\[f_{m_{k}}=(I-\lambda^{-1}K)f_{m_{k}}+\lambda^{-1}Kf_{m_{k}}\to f,\]  
because \((I-\lambda^{-1}K)f_{m_{k}}\to 0\) due to the properties of the N-iteration (Proposition 1). The continuity of \(I-\lambda^{-1}K\) implies that \(0=(I-\lambda^{-1}K)f\) and \(f\in Fix(\lambda^{-1}K)\). The fact that \(\lim_{n\to\infty}||f_{n}-f||\) exists (see Proposition 1) implies that the sequence of N-iterates \((f_{n})\) converges strongly to the fixed point \(f\) (that is to say, to an eigenvector of \(K\) associated with \(\lambda\)). **Corollary 1**: _Let \(\lambda\) be a non-null eigenvalue of the real or complex matrix \(A\). If, for some subordinate matrix norm, \(||A||\leq|\lambda|\), then the N-iteration applied to \(\lambda^{-1}A\) is convergent to an eigenvector associated with \(\lambda\)._  
This is a trivial consequence of Theorem 1 considering that, in finite dimensional spaces, all the linear operators are compact. **Example 6**: _The matrix_  
\[A= \begin{pmatrix}4&5\\ 6&5\end{pmatrix}