# Is there an infinite dimensional $V$ such that every linear operator on $V$ is a direct sum of a nilpotent and an invertible operator?

I know that if $V$ is a finite-dimensional vector space and $T$ is a linear operator on $V$, then $T$ is a direct sum of a nilpotent operator and an invertible operator.

My question is: Is there an infinite-dimensional vector space $V$ over a field $\mathbb{F}$ such that every linear operator $T$ on $V$ can be written as a direct sum of a nilpotent linear operator and an invertible linear operator?

I mean "There are subspaces such that $V=M\oplus N$ where $T$ is invertible on $M$ and $T$ is nilpotent on $N$"?

-
Do you mean "semi-simple" instead of invertible? Because the zero transformation can't ever be written as a sum of an invertible and nilpotent operator (except for dimension 0). – Bill Cook Feb 9 '12 at 20:54
The zero transformation is nilpotent. I will write more in the text. Please, just a second. – spohreis Feb 9 '12 at 21:03
Yes. Zero is nilpotent. But you can't write $0=I+N$ where $I$ is invertible and $N$ is nilpotent. Do you mean "There are subspaces such that $V = M \oplus N$ where $T$ is invertible on $M$ and $T$ is nilpotent on $N$"? I'm not sure I'm clear on what you're trying to ask. – Bill Cook Feb 9 '12 at 21:30
@BillCook: Yes! I will add it to my question. Thank you for point ir out. – spohreis Feb 9 '12 at 21:39

No (at least assuming AC). Let $V$ be an infinite dimensional vector space, let $B$ be a basis for $V$, let $b\in B$, and let $f:B\to B\setminus\{b\}$ be a bijection. Let $T:V\to V$ be the unique linear transformation such that $T|_B=f$. Since $T$ is injective, its restriction to every subspace is injective, so the only subspace on which $T$ is nilpotent is $\{0\}$. But $T$ is not invertible, because $b$ is not in its range.

All that was needed for this example is that there exists an injective but not surjective linear operator on $V$. Similarly, a surjective but not injective operator has no nilpotent summand on a nontrivial subspace, and is not invertible.

-