Find the basis for the eigenspace. Can it be the zero vector? I am trying to obtain a basis for an eigenspace given the standard matrix of a linear operator over a space. I have done all of the work. I just need to confirm my results or find my mistake. 
A=[F]=
   \begin{array}{ccc}
   3 & 2 & 1 \\
   0 & 2 & 4 \\
   0 & 0 & 4
  \end{array} 
After finding $|\lambda I - A|$ I get that the eigenvalues are $\lambda_{1}=2$, $\lambda_{2}=3$ and $\lambda_{3}=4$. I am having a problem with $\lambda=4$. When I compute $4I-A$, the computation yields that there is no basis for the nullspace, does this mean that there is no basis for this eigenspace? 
4I-A=
  \begin{array}{ccc}
   1 & -2 & -1 \\
   0 & 2 & -4 \\
   0 & 0 & -4
  \end{array} 
So then upon row reducing to the canonical form $C$, I obtain the identity matrix. 
I_3=
  \begin{array}{ccc}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1
  \end{array} 
This says that $x=y=z=0$.
So, is this the basis for the eigenspace? The basis for the nullspace does not exist since the only vector in the basis of the nullspace is the zero vector.
 A: No. A basis is a linearly in-dependent set. And the set consisting of the zero vector is de-pendent, since there is a nontrivial solution to $c\vec{0}=\vec{0}$.
If a space only contains the zero vector, the empty set is a basis for it. This is consistent with interpreting an empty sum as zero.

But with respect to your computation, you are just not computing $4I-A$ correctly. Look again at the lower right entry.
A: Since nobody seems to have computed it correctly yet, let me just say that the nullspace of $4I-A$ is generated by the vector $v=(5,2,1)$; indeed $Av=(20,8,4)=4v$.
A: It is a theorem that $\det(B) = 0 \iff B$ is singular.  So the very fact that an eigenvalue exists implies that the matrix $A-\lambda I$ is singular: we have found a $\lambda$ such that $\det(A-\lambda I) = 0$.
Therefore, in this case, $\operatorname{Rank}(A-4I) < 3$, and so $A - 4I$ must have at least one row of zeros in its reduced row echelon form.  Note that this also implies a nontrivial nullspace:  The rank-nullity theorem tells us that $\operatorname{Rank}(B) + \dim( \operatorname{Null(B)}) = n$ for any $n \times n$ matrix $B$.  
In other words, the eigenspace for a given eigenvalue must have at least one nontrivial basis vector.
Long story short, you've made an error (highlighted by Omnnomnomnom in his answer).
