Endomorphism Ring: $f$ zero divisor $\iff$ $\dim Im f< \dim V$

Question

Let $V$ be finite dimensional and $f \in \mathcal{L}(V,V)$, show that:

$f$ zero divisor $\iff$ $\dim Im f< \dim V$

My ideas:

$\Leftarrow$ Let $B$ be a basis of $V$. Since $0 < \dim V- \dim Im f=\dim \ker f$, $f$ is not injective and therefore we can find a $v\neq 0$ such that $f(v)=0$, $\{v\}$ is linear independent in $V$ we can find a basis $B_{v}$ in $V$, and define $g \in \mathcal{L}(V,V)$ such that $g(v)=v$ and $g(b)=0$ , for all $b \in B_{v}-\{v\}$ It follows that $f\circ g(v)=0$

$1.$ Question I have that this is the case for $f\circ g=0$ but surely I now need to find $h$ such that $h \circ f=0$, how do I do this?

for the case "$\Rightarrow$" I am struggling.

Answer 1

The idea is good, but it can be improved.

Suppose $\dim\operatorname{Im}f<\dim V$. Then there exists $v\ne0$ such that $f(v)=0$. Complete $v$ to a basis $\{v_1=v,v_2,\dots,v_n\}$ of $V$ and define $g\colon V\to V$ such that $g(v_1)=v_1$, and $g(v_i)=0$ for $i=2,\dots,n$. Then clearly $g\ne0$ and $f\circ g=0$.

How can we find $h\colon V\to V$ such that $h\ne0$ and $h\circ f=0$? Just take a basis $\{w_1,\dots,w_m\}$ of $\operatorname{Im}f$ (it could be empty if $f$ is the zero endomorphism) and complete it to a basis $\{w_1,\dots,w_m,w_{m+1},\dots,w_n\}$ of $V$. Define $h(w_i)=0$ if $1\le i\le m$ and $h(w_i)=w_i$ if $m<i\le n$. Since $m<n$, you have $h\ne 0$.

The converse is clear: if $f$ is a zero divisor, then it cannot be invertible, so $\ker f\ne\{0\}$ and therefore $\dim\operatorname{Im}f<\dim V$.

Answer 2

If $f$ doesn't have full rank, you can find a basis of its image and extend it to a basis of $V$. Then define a map $h : V\to V$ that is zero on the basis elements of the image of $f$ and the identity on the remaining ones. Then $hf=0$ but $h$ is not zero.

Answer 3

Suppose that there is a $g\in\mathcal{L}(V,V)$ such that $f\circ g=0$. If $\dim\operatorname{im}f=\dim V$, then $f$ would have an inverse, and therefore\begin{align}g&=\operatorname{Id}\circ g\\&=(f^{-1}\circ f)\circ g\\&=f^{-1}\circ(f\circ g)\\&=f^{-1}\circ0\\&=0.\end{align}