Q:

Questins on Formulae for Eigenvalues & Eigenvectors of any 2 by 2 Matrix

Let $A =  \begin{bmatrix}
    a & b \\
    c & d \\ 
    \end{bmatrix}$. Then $\det(A - \lambda I) = 0 \implies \lambda_{\pm}  \frac{\text{Trace} \pm \sqrt{\text{Trace}^2 - 4\det}}{2} = \frac{(a + d) \pm \sqrt{(a - d)^2 + 4bc}}{2}$.
Case 1: If $c \neq 0$, then $\mathbf{v_1} =\begin{bmatrix}
    \lambda_+ - d \\ c
    \end{bmatrix}^T $ and $\mathbf{v_2} = \begin{bmatrix}
    \lambda_- - d \\ c
    \end{bmatrix}^T $
Case 2: If $b \neq 0$, then $\mathbf{v_1} = \begin{bmatrix}
    b \\ \lambda_+ - a 
    \end{bmatrix} $ and $\mathbf{v_2} = \begin{bmatrix}
    b \\ \lambda_{-} - a 
    \end{bmatrix} $
Case 3: If $b = c = 0$, then $\mathbf{v_1 = e_1}^T$ and $\mathbf{v_2 = e_2}^T $ 

$\Large{{2.}}$ After calculating the eigenvalues, how would you analyse $(A - \lambda I)\mathbf{x}  = \mathbf{0} $ in general?
Then how and why would you determine to take cases? 
$\Large{{3.}}$ How and why only $3$ cases? 
What about the case $a, b, c, d \neq 0$? Then $(A - \lambda I)\mathbf{x}  = \mathbf{0} $
$\implies \left[
      \begin{array}{cc|c}
        a-\lambda & b & 0 \\
    c & d-\lambda & 0\\
      \end{array}
    \right]$. $\det(A - \lambda I) = 0 \iff$ $A$ singular $\iff$ Rows lin-dependent. Thus, for each $\lambda$,
$\mathbf{v} = \begin{bmatrix}
    \color{#FF4F00}{\LARGE{-}}\text{ entry in 2nd column of either row } \\ \text{ entry in first column of either row }
    \end{bmatrix} : \begin{bmatrix}
    \color{#FF4F00}{-}(d - \lambda) \\ c
    \end{bmatrix} $ or $\begin{bmatrix}
    \color{#FF4F00}{-}b \\ a - \lambda
    \end{bmatrix} $.
Moreover, what about the case where $b = 0$ ? Then $a - \lambda$ is a pivot so $rank(A - \lambda I) = 1 \text{ or } 2$?

A:

Let's compute the eigenvalues of $A_{2\times 2}$ in the general case: the characteristic polynomial is
$$
  f_A(\lambda)
= \det\left[
  \begin{array}{cc}
  \lambda - a & -b \\
  -c          & \lambda - d
  \end{array}
  \right]
%= (\lambda - a)(\lambda - d) - bc
= \lambda^2 -(a+d)\lambda +ad- bc,
$$
which has roots
$$
  \lambda
= \frac{a+d \pm \sqrt{(a  +d)^2 - 4(ad-bc)}}{2}
= \frac{a+d \pm \sqrt{(a-d)^2 \color{red} + 4bc}}{2}.
$$
For the second question, I'm not sure exactly what you are asking, but after finding an eigenvalue $\lambda$ one generally analyzes $(A - \lambda I) {\bf x} = {\bf 0}$ in order to find elements of the eigenspace corresponding to $\lambda$, and this is done by Gaussian elimination. Illustrating with the matrix $A$ given above,
$$
     [A - \lambda I]
=    \left[
     \begin{array}{cc}
     -1 & 1/2 \\
     -2 & 1 \\
     \end{array}
     \right]
\sim \left[
     \begin{array}{cc}
     1 & -1/2 \\
     0 &  0 \\
     \end{array}
     \right];
$$
the first row corresponds to the equation $x_1-\frac{1}{2}x_2=0$, and so the eigenvectors corresponding to $\lambda=2$ are of the form ${\bf x} = t[1, 2]^\top$, $t \in \mathbb{R}$. This process generalizes to any finite dimensional matrix, and your solution set may contain more basic solutions (i.e. your eigenspace may be of higher dimension than $1$, as happened here). I hope this answers your second question.
I'm a little confused about the formulae given for the eigenvectors in your post, because I can't tell if you mean for the vectors in case 1 to be left eigenvectors or not (you superscripted them with ${}^T$). To explain why the cases don't seem to exhaust all possibilities, think along these lines: either both $c$ and $b$ are equal to $0$ (case 3), or at least one of the is not (cases 1 and 2 together).
Lets try to find a general formula for the (right) eigenvectors, using the evalues we found above: note that in case 3
$$
  [A - \lambda I]
= \left[
  \begin{array}{cc}
  a - \lambda & 0 \\
  0           & d - \lambda \\
  \end{array}
  \right];
$$
now, since you know that $A - \lambda_1 I$ is not invertible, it must be that at least one of these rows is a zero row. If it is the second row then ${\bf e}_1$ is a solution (and hence an evector for $\lambda$), if it is the first row then ${\bf e}_2$ is the solution and evector. If both rows are zero ($\lambda$ was a double root of $f_A$) then both are evectors. If one of the rows was non-zero in $[A - \lambda_1I]$, the opposite row will be non-zero in $[A - \lambda_2 I]$ and yield the opposite evector. This handles case 3.
If $c \neq 0$ then since
\begin{align}
      [A - \lambda I]
      \left[
      \begin{array}{c}
      \lambda - d \\
      c             \\
      \end{array}
      \right]
&=    \left[
      \begin{array}{c}
      (a-\lambda)(\lambda-d) + bc \\
      c(\lambda-d) + (d-\lambda)c \\ 
      \end{array}
      \right] \\
&=    \left[
      \begin{array}{c}
      (a-\lambda)(\lambda-d) + bc \\
      0 \\ 
      \end{array}
      \right],
\end{align}
case 1 can be verified by showing that 
$(a-\lambda_+)(\lambda_+-d) + bc = (a-\lambda_-)(\lambda_--d) + bc=0$ when $c\neq 0$. Dealing with $\lambda_+$ you find
\begin{align}
   (a-\lambda_+)(\lambda_+-d) + bc
&= \frac{1}{4}(a-d - \sqrt{(a-d)^2 + 4bc})(a-d + \sqrt{(a-d)^2 + 4bc}) + bc \\
&= \frac{1}{4}((a-d)^2 - (a-d)^2 - 4bc) + bc \\
&= 0.
\end{align}
The computation for $\lambda_-$ is almost identical. So this handles case 1, and case 2 is similar.
I hope that clears everything up! 

