A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}{.8}&0\\0&{.2}\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}{.4}\\{.1}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.32}\\{.02}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.256}\\{.004}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.2048}\\{.0008}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.16384}\\{.00016}\end{aligned}} \right)\)

From the above conclusion, \({A^k}{\bf{x}}\) is approximate \(.8\).

Conclusion: If \({\bf{x}} \ne 0\) and all the eigenvalues of the matrix \(A\) are less than \(1\) in magnitude then \({A^k}{\bf{x}}\) is approximately an eigenvector for the larger \(k\).

Consider \(A = \left( {\begin{aligned}{ {20}{c}}1&0\\0&{.8}\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}{.5}\\{.4}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.32}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.256}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.2048}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.16384}\end{aligned}} \right)\)

From the above conclusion \({A^k}{\bf{x}}\) is approximate \(\left( {\begin{aligned}{ {20}{c}}{.5}\\0\end{aligned}} \right)\).

Conclusion: If the strictly dominant eigenvalue of \(A\) is \(1\), and if \(x\) has a component in the direction of the corresponding eigenvector, then \({A^k}{\bf{x}}\) will converge to a multiple of that eigenvector.

Consider \(A = \left( {\begin{aligned}{ {20}{c}}8&0\\0&2\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}4\\1\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{32}\\2\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{256}\\4\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{2048}\\8\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{16384}\\{16}\end{aligned}} \right)\)

From the above conclusion distance of \({A^k}{\bf{x}}\) distance of from either eigenvector of \(A\) is increasing rapidly as \(k\) increases

Conclusion: If the eigenvalues of \(A\) are all greater than \(1\) in magnitude, and if \(x\) is not an eigenvector, then the distance from \({A^k}{\bf{x}}\) to the nearest eigenvector will increase as \(k \to \infty \).