Why doesn't the general formula $(A - \lambda I) \eta = v$ for Eigenvectors of repeated eigenvalues work? Given the following matrix
\begin{bmatrix}
    1 & 0 & 0 & 0 \\
    0 & -1 & 1 & 0  \\
    1 & -1 & 0 & 1 \\
    1 & -1 & -1 & 2 
\end{bmatrix}
Which has eigenvalues $\lambda = 0, 1$, each with multiplicity of 2.
Proceed to find eigenvectors for $\lambda = 1$.
This gives the following "characteristic matrix",
\begin{bmatrix}
    0 & 0 & 0 & 0 \\
    0 & -2 & 1 & 0  \\
    1 & -1 & -1 & 1 \\
    1 & -1 & -1 & 1 
\end{bmatrix}
The formula $(A-\lambda I) v = 0$ gives:
\begin{align*}
-2v_2 + v_3 &= 0\\
v_1 -v_2 -v_3 + v_4 &= 0
\end{align*}
By choosing the appropriate pair of $v_i$, one can obtain the eigenvectors corresponding to $\lambda = 1$, which are:
$v = (3,1,2,0)$ and $\eta = (1,0,0,-1)$.

My question is that: Suppose instead of playing the guessing game, I applied the general formula to obtain the eigenvector:
$$(A - \lambda I) \eta = v$$
But this does not work due to the $(0,0,0,0)$ row - algebraically speaking.
I thought that equation was supposed to be the general form, which should work for all cases. Could someone please explain why the formula does not work - not why there are only 2 eigenvalues and eigenvectors?
Also, if you down vote this, please leave an explanation.
 A: I may be able to clear this up for you.  I looked at the website that you cited.  They are trying to find solutions to the system
$$x'(t) = Ax(t) $$
where $A$ is a $2 ×2$ linear operator with a repeated eigenvalue. However, consider the possible Jordan forms that $A$ can have; there are only two:
$$J_A = \begin{bmatrix}\lambda & 0\\0 & \lambda\end{bmatrix}  or \:\begin{bmatrix}\lambda & \color{red}1\\0 & \lambda\end{bmatrix}.$$  That's it.
So if $J_A$ is the first, then $A$ is not only $similar$ to $J_A$, $A$ must be $EQUAL$ to $J_A$ (this is very easy to show).  If that is the case, then your ODE is now
$$x'(t) = \begin{bmatrix}\lambda & 0\\0 & \lambda\end{bmatrix}x(t) = \lambda Ix(t) = \lambda x(t); $$
i.e reduced to a much more basic equation.  The only other option is the second, and we can see from the presence of the "1" above the $\lambda$ instead of a "0" that the dimension of the nullspace of $A - \lambda I $ is 1, meaning there must be a generalized eigenvector.  To clarify, if $A$ is $2×2$ with a repeated eigenvalue and is not a scalar multiple of the identity, then once we find $v$ such that $Av = \lambda v$, then we are $GUARANTEED$  a non-zero (obviously) vector $u$ such that $(A - \lambda I)u = v$.
But in higher dimensions this is not always the case, and we can't just assume it is true.   Let's look now at your $4×4$ matrix.  It has the following Jordan decomposition:
$$ \begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 1 & 0 \\1 & -1 & 0 & 1 \\1 & -1 & -1 & 2\end{bmatrix}
=  U
\begin{bmatrix}0 & \color{red}1 & 0 & 0\\0 & 0 & 0 & 0 \\0 & 0 & 1 & \color {red} 0 \\0 & 0 & 0 & 1\end{bmatrix}
U^{-1}$$
Notice the block of two "1"s on the diagonal.  The "0" that is above one and to the right of the other indicates that the eigenspace associated with $\lambda = 1$  is two-dimensional; i.e. there are only genuine eigenvectors for the eigenvalue 1.  This means that once you solve for $Av = v$, then $v$ will be expressible as:
$$v = se_1 + te_2 $$
where $e_1$ and $e_2$ serve as a basis for the eigenspace, and there will be no $u$ such that $(A-I)u = v$.
Additionally, we can see that the "1" in the corner of the block with "0"s on the diagonal means that the kernel of $A -0I = A$ is actually one-dimensional, so once you find $THAT$ corresponding eigenvector $v$ with $Av = 0$, there $WILL$ be a non-zero $u$ such that $(A-0I)u=v$.
Hope this helps.
A: True or "first order" eigenvectors satisfying $Av = \lambda v$ will live in an undiscernable space in the sense that you can choose vectors freely and they will always map on themselves and never accidentally "cross-map" on the other eigenvalue of the "different" geometry. It is only if the algebraic and geometric multiplicity of eigenvalue differ that you can get "fused" or "cross"-mappings in different senses.
An eigenvalue $0$ really does not help make any cross-mapping interpretation to make sense because the whole sub-space will necessarily be nilpotent.
But just to show how to do it, calculate the eigenvector $[0,1,1,1]^T$, ok?
Now try and solve $(A-0*I)w = [0,1,1,1]^T$, you will find $w = [0,-2,1,1]^T$ and if you multiply $Aw$ you will get $3[0,1,1,1]^T$
So the complicated space maps $[0,-2,1,1]$ to 0 times itself + 3 times $[0,1,1,1]^T$ and in turn the $[0,1,1,1]^T$ maps to itself but shrinks to 0 due to the eigenvalue being $0$.

For the term cross this is what I mean:
$${\bf M} = \left[\begin{array}{cc}-1&2\\-2&1\end{array}\right], {\bf S} = \left[\begin{array}{rr}1&1\\1&-1\end{array}\right], {\bf S}^{-1}\bf {MS} = \left[\begin{array}{rr}0&-3\\1&0\end{array}\right]$$
Multiplication with $\bf M$ causes the vectors $[1,1]^T$ and $[1,-1]^T$ to multiply onto each other one being multiplied by $-3$ in the process and the other by $1$, if you draw it with images with elements as dots and the mapping like arrows, it would look like a cross.

For the term fuse I mean for example this matrix
$$\left[\begin{array}{rr}1&1\\0&1\end{array}\right]$$
When multiplying a vector with this matrix the first element sums or fuses the two values in the vector. So that the vector $[1,0]$ after transformation contains the sum before the transformation.
