What are the eigenvectors of 4 times the identity matrix? What are the eigenvectors of the identity matrix
$
 I=
  \left[ {\begin{array}{cc}
   1 & 0 & 0 \\       0 & 1 & 0 \\    0 & 0 & 1 \\  \end{array} } \right]
$
 A: From
$$I\vec v=\vec v$$
you conclude that any vector is an Eigenvector of $I$, with Eigenvalue $1$. (This is explained by the fact that the Eigenvalue has multiplicity $d$ in $d$ dimensions.)
A: When you want to find the eigenvectors and eigenvalues, the first thing we do is to diagonalize the matrix. Hey! That has already been done! Now we can just read of the eigenvalues (they are the diagonal entries) and a set of eigenvectors (they are the three column vectors - let's call them $e_1,e_2,e_3$). 
To give context, let's see if the matrix in its current form actually satisfy the eigenequation $A v = \lambda v$. It does:
$$\mathbb{I} \; e_n = 1 \cdot e_n$$
where $\mathbb{I}$ is the identity matrix (it is called the "identity" matrix because it always returns the vector it is working on untouched - try it with any vector other than then null-vector!). Check this for yourself and let me know if it still doesn't make sense. 
A: Follow the definition. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. 
Substitute one eigenvalue λ into the equation Ax = λx or, equivalently, into (A − λI)x = 0 and solve for x; the resulting nonzero solutions form the set of eigenvectors of A corresponding to the selected eigenvalue. This process is then repeated for each of the remaining eigenvalues.
