# 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 $T(\vec v)=S_T' (\vec v) + S_T''(\vec v)+... \forall \vec v \in V$ ?

• I note that the problem is solved by N. J. Lord in the paper, Matrices as Sums of Invertible Matrices, published in Mathematics Magazine in February, 1987. The paper may be available at maa.org/sites/default/files/N06425._J._Lord.pdf A generalization is discussed at MO, mathoverflow.net/questions/141382/… – Gerry Myerson Sep 30 '14 at 6:05
• @GerryMyerson: Can we not complete the solution by going in the direction of non-zero eigenvalues as suggested by Robert Israel ? – Souvik Dey Sep 30 '14 at 6:08
• Well, you could have a look first at the Lord paper, to see how he does it. The problem with Robert's hint, as we have seen, is that over a finite field there may not be any $t$ that isn't an eigenvalue of $T$. – Gerry Myerson Sep 30 '14 at 6:11

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).

• Indeed, this shows $T$ is a sum of a rather small number of invertible operators. – Gerry Myerson Sep 28 '14 at 9:04
• Nice. What happens if the base field is finite, though? – k.stm Sep 28 '14 at 9:44
• @Robert Israel : Ok , so then $T(\vec v)=t\vec v + \big (T(\vec v) -t\vec v \big)$ ; but can we ensure that there is a non-zero scalar which is not an eigenvalue ? – Souvik Dey Sep 28 '14 at 13:09
• That would depend, Souvik, on the underlying field. – Gerry Myerson Sep 29 '14 at 0:22
• @GerryMyerson , Robert Israel : So then how could we solve the problem if every non-zero scalar is an eigenvalue ?? – Souvik Dey Sep 29 '14 at 4:26

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)

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.