How to show that a linear subspace is a direct sum of two vectors invariant by an application? Suppose
$$
A= \begin{bmatrix}
1  & 1 & -1\\ 
 -2& -1 & 2\\ 
 0& 1 & 0 
\end{bmatrix}
$$
and $B=\text{span}\{(0,1,1),(1,1,3)\}$. 
In order to show that $B$ is a sum of two invariant vectors by A, is it enough for me to show that first the vectors $(0,1,1), (1,1,3)$ are linearly independent and that they are invariant by $A$? 
 A: There seems to be lots of confusion in the question itself. 


*

*An invariant subspace of a linear mapping $T : V → V$ from some vector space $V$ to itself is a subspace $W$ of $V$ that is preserved by $T$; that is, $T(W) \subset W$.
If you write
$$
W=\color{blue}{\textrm{span}}\{u,v\}
$$
where $u=(0,1,1)$ and $v=(1,1,3)$, then one can ask the following question:

Is $W$ an invariant subspace of the linear mapping defined by the matrix $A$?


*Since you are using row vectors here, you might want to be careful how you define the linear map using via the matrix $A$. 

*A "vector" is an element of a vector space, it is NOT even a vector space and thus cannot be called a subspace of some vector space. 

*Direct sum of vector spaces are different from the sum of two vectors. 

*One only says "eigenvectors" and "invariant subspaces". There is no such a thing called "invariant vector". If a vector is "unchanged" with respect to a linear map, say, in your case $A(x)=x$, then it would be called a "fix point" of $A$, or in the setting or linear algebra, an eigenvector with respect to the eigenvalue $1$. 


With the above clarification of your question, can you go on with your question now?
