I am trying to understand the method of finding eigen vectors in the case of repeated eigen values. My question is based on section 3.5.2 of this link. In the first example (Example 3.5.4),
$A = \begin{bmatrix}2&0\\0&2\end{bmatrix}$
Here, the $\lambda = 2$ (repeated eigen value). If $(A - \lambda I)$ is calculated, it is zero matrix of dimension 2. So, the geometric multiplicity is 2, which means there must be two linearly independent eigen vectors.
1) Am I correct in understanding that these can be any two linearly independent vectors since $(A - \lambda I)$ is a zero matrix? Or is there a reason for picking $v_1 = (1, 0)$ and $v_2 = (0, 1)$ as shown in the link?
Now, consider the second example (Example 3.5.5):
$A = \begin{bmatrix}5&1\\-4&1\end{bmatrix}$
In this case, $\lambda = 3$ (repeated eigen value) and
$A - \lambda I = \begin{bmatrix}2&1\\-4&-2\end{bmatrix}$
Here geometric multiplicity of $\lambda$ is 1. So, there is only one linearly independent eigen vector.
2) What is the idea behind using $(A - \lambda I)v_2 = v_1$ to find second eigen vector?
3) Is this technique used only when geometric multiplicity is less than algebraic multiplicity? Otherwise, do we just use logic to find all independent eigen vectors as we did in the first example?
4) Since the matrix is not diagonalizable, is the idea to minimize the error, rather than solve the system of equations? I am trying to understand the need to find a second eigen vector in practical situations.