Does $\ker T\cap {\rm Im}\,T=\{0\}$ imply $V=\ker T\oplus{\rm Im}\,T$? Let $T: V\rightarrow V$ be a linear operator of the vector space $V$.
We write $V=U\oplus W$, for subspaces $U,W$ of $V$, if $U\cap W=\{0\}$ and $V=U+W$.
If we assume $\dim V<\infty$, then by the rank-nullity theorem, $\ker T\cap {\rm Im}\,T=\{0\}$ implies $V=\ker T\oplus {\rm Im}\,T$.

However, my question is about the case $\dim V$ is infinite. Is it still true?  What if $T$ has a minimal polynomial?

Thanks.
 A: Consider the shift operator $s$, defined on $\text{Vect}(e_i, i\in\mathbb{N})$, where $s(e_n)=e_{n+1}$ for $n\in\mathbb{N}$.  Note that $\ker(s)=0$ but $s$ is not surjective.
A: Let $\mathbb{K}$ be the base field.  If $T:V\to V$ is such that $\ker(T)\cap\text{im}(T)=0$ and there exists $p(X)\in\mathbb{K}[X]$ such that $p(T)=0$, then $$V=\ker(T)\oplus\text{im}(T)\,.$$
By choosing $p(X)$ to be the monic polynomial of the lowest possible degree, we may assume that $0$ is a simple root of $p(X)$ (this is due to the assumption that $\ker(T)\cap\text{im}(T)=0$, and if the minimal polynomial of $T$ is not divisible by $X$, which is possible, then we simply multiply the minimal polynomial of $T$ by $X$).  That is,
$$p(X)=X^n+a_{n-1}X^{n-1}+a_{n-2}X^{n-2}+\ldots+a_2X^2+a_1X$$
for some $a_1,a_2,\ldots,a_{n-2},a_{n-1}\in\mathbb{K}$ with $a_1\neq 0$.
Write $q(X):=X^{n-1}+a_{n-1}X^{n-2}+a_{n-2}X^{n-3}+\ldots+a_2X+a_1$.  Note that
$$1=\frac{1}{a_1}\,q(X)+r(X)\,X\,,$$
where
$$r(X):=-\frac{1}{a_1}\,X^{n-2}-\frac{a_{n-1}}{a_1}\,X^{n-3}-\frac{a_{n-2}}{a_1}\,X^{n-4}-\ldots-\frac{a_3}{a_1}\,X-\frac{a_2}{a_1}\,.$$
Therefore,
$$\text{id}_V=\frac{1}{a_1}\,q(T)+r(T)\,T\,.$$
Fix $v\in V$.  We get
$$v=\text{id}_V(v)=\left(\frac{1}{a_1}\,q(T)+r(T)\,T\right)v=\frac{1}{a_1}\,q(T)v+r(T)\,Tv\,.$$
Observe that $q(T)v\in \ker(T)$ and $Tv\in\ker\big(q(T)\big)$ (as $X\,q(X)=q(X)\,X=p(X)$ is the minimal polynomial of $T$).  This implies
$$V=\ker(T)\oplus\ker\big(q(T)\big)\,.$$
We want to prove that
$$\text{im}(T)=\ker\big(q(T)\big)\,.$$
The direction $\text{im}(T)\subseteq\ker\big(q(T)\big)$ is clear because $q(X)\,X=p(X)$.  We shall prove the reversed inclusion.  Suppose that $v\in\ker\big(q(T)\big)$.  Thus,
$$T^{n-1}v+a_{n-1}\,T^{n-2}v+a_{n-2}\,T^{n-3}v+\ldots+a_2Tv+a_1v=0\,.$$
This gives
$$v=T\left(-\frac{1}{a_1}\,T^{n-2}v-\frac{a_{n-1}}{a_1}\,T^{n-3}v-\frac{a_{n-2}}{a_1}\,T^{n-4}v-\ldots-\frac{a_2}{a_1}\,v\right)\in \text{im}(T)\,.$$
