Linear Algebra/Topic: The Method of Powers/Solutions

Solutions

Problem 1

Use ten iterations to estimate the largest eigenvalue of these matrices, starting from the vector with components $1$ and $2$. Compare the answer with the one obtained by solving the characteristic equation.

1. ${\begin{pmatrix}1&5\\0&4\end{pmatrix}}$
2. ${\begin{pmatrix}3&2\\-1&0\end{pmatrix}}$

Answer

1. The largest eigenvalue is $4$.
2. The largest eigenvalue is $2$.

Problem 2

Redo the prior exercise by iterating until $|{\vec {v}}_{k}|-|{\vec {v}}_{k-1}|$ has absolute value less than $0.01$ At each step, normalize by dividing each vector by its length. How many iterations are required? Are the answers significantly different?

Problem 3

Use ten iterations to estimate the largest eigenvalue of these matrices, starting from the vector with components $1$, $2$, and $3$. Compare the answer with the one obtained by solving the characteristic equation.

1. ${\begin{pmatrix}4&0&1\\-2&1&0\\-2&0&1\end{pmatrix}}$
2. ${\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}}$

Answer

1. The largest eigenvalue is $3$.
2. The largest eigenvalue is $-3$.

Problem 4

Redo the prior exercise by iterating until $|{\vec {v}}_{k}|-|{\vec {v}}_{k-1}|$ has absolute value less than $0.01$. At each step, normalize by dividing each vector by its length. How many iterations does it take? Are the answers significantly different?

Problem 5

What happens if $c_{1}=0$? That is, what happens if the initial vector does not to have any component in the direction of the relevant eigenvector?

Answer

In theory, this method would produce $\lambda _{2}$. In practice, however, rounding errors in the computation introduce components in the direction of ${\vec {v}}_{1}$, and so the method will still produce $\lambda _{1}$, although it may take somewhat longer than it would have taken with a more fortunate choice of initial vector.

Problem 6

How can the method of powers be adopted to find the smallest eigenvalue?

Answer

Instead of using ${\vec {v}}_{k}=T{\vec {v}}_{k-1}$, use $T^{-1}{\vec {v}}_{k}={\vec {v}}_{k-1}$.