In diagonalisation and, I think, generally speaking, why are we able to change lines of $A-\lambda I$ and not those of $A$?
Is it because when the search for the eigen values of $A$ we assume that we are solving a system of linear eauations where $\lambda$ is the unknown?
Furthermore, why doing such an operation is not correct and gives 5 eigen-values: \begin{bmatrix} 1 & -\lambda & 1 & \\ 0 & 1-\lambda^2 & \lambda \\ 0 & 1 & -\lambda \\ \end{bmatrix}
$$\Leftrightarrow \begin{bmatrix} 1 & -\lambda & 1 & \\ 0 & 1-\lambda^2 & \lambda \\ 0 & 0 & -\lambda(1-\lambda^2)-\lambda \\ \end{bmatrix}$$
Why do we have to do the following one, which allow us to have 3 eigen values:
\begin{bmatrix} 1 & -\lambda & 1 & \\ 0 & 1 & -\lambda \\ 0 & 1-\lambda^2 & \lambda \\ \end{bmatrix}
When you find eigenvalues by row reducing $A - \lambda I$, you're just manipulating coefficients of a system of linear equations by conventional elementary row operations. If you operate on $A$ instead of on $A - \lambda I$, you're solving a different set of equations (which happen to be those giving the null space of $A$).
Not sure what your proposed operation is that gives five solutions, but it's not a sequence of elementary row operations.