How to prove that every linear operator on a finite dimensional vector space is a sum of invertible linear operators Let $V$ be a finite dimensional vector space , then how do we prove that for every linear operator $T$ on $V$ , there exist invertible linear operators $S_T' , S_T'',...$ such that $T(\vec v)=S_T' (\vec v) + S_T''(\vec v)+... \forall \vec v \in V$ ?
 A: Hint: if $t \ne 0$ is not an eigenvalue of $T$, then ...
EDIT: In the case of dimension $1$ over the field $GF(2)$ with two elements, 
there is only one invertible linear operator ($1$), and it is not the sum of two invertible linear operators (though it is the sum of three).
A: HINT: If the set $\det A \ne 0$ was contained in a hyperplane $l(A) = 0$ then the function $l \cdot \det$ would be identical $0$. 
( works for infinite fields)
A: Recall that every matrix can be expressed as a sum of a diagonal matrix and an invertible matrix.
Also every diagonal matrix can be expressed as a sum of 2 invertible matrices.
Now since the Endomorphism ring End(V) of a finite dimensional vector space V is an n*n matrix ring, your result is proved.
Infact we can say that further from the results of Wolfson {An ideal theoretic characterization of the ring of all linear transformations} and Zelinksy {Every Linear Transformation is Sum of Non- singular Ones} that if V is not one dimensional over a field of 2 elements then every element in End(V) is a sum of 2 invertible linear transformations.
