Using Power Method to get Eigen-vector?

Question

Edit: please take a look at the comments section

I learnt that power method converges to the Eigen vector which suits the biggest Eigen value (in absolute value)

Now suppose I have the following Eigen values for matrix $A$ $\{3,1,-3\}$

I can get $v_1$ (which matches $3$) Simply by using power method on $A+I$ -Since the new Eigen values will be $\{4,2,-2\}$ and $v_1$ will have the biggest Eigen value-

Similary, to get $v_3$ I will use power method on $A-I$.

But what about $v_2$, how may I get it?

I thought about making the Eigen values equal to $\{1/3, 1/1, -1/3\}$ But how may I do that?

For example, $A$ is:

Answer 1

There are two approaches you can try.

1. Inverse Power Method: This method relies on the usage of $A^{-1}$ for a power iteration, ie you solve a linear system with $A$ repeatedly. You can easily see, that this matrix exists as there are no EVs equal to $0$. Let $\lambda, v$ be an eigenpair of $A$, with $Av=v\lambda$. Then $A^{-1}v = v \lambda^{-1}$ which is easy to prove. This gives that $\lambda^{-1},v$ is an eigenpair of $A^{-1}$. As your matrix $A$ has EVs $(3,1, -3)$ you inverted matrix $A^{-1}$ will have the EVs $(1/3, 1, -1/3)$, where $1$ is the EV with the largest absolute value. This is the limit of the iteration.

2. Power method with projection: After you obtained the eigenvectors for $3$ and $-3$ by shifted power iteration you can find the last eigenvector by using the power iteration and subtract any projection along the already known eigenvectors. This is best used for normal matrices with $AA^T = A^TA$ (for example symmetric or skew-symmetric matrices).