When does the kernel of a function equal the image? When does the kernel of a function equal the image?
 Thanks in advance  
 A: If you are talking about linear transformations of $\mathbb{R}^n$ to itself, the kernel is the space orthogonal to the rows.  The image however is the column space.  So your kernel equals the image when the rows are all orthogonal with all the columns and the rank is $\frac{n}{2}$.
A: Let $f$ be a function which takes $x$ as argument such as $f(x)=y$.
If $x \in \text{Ker} f$ then $f(x)=0 \Longrightarrow f(0)=0$.
If $y \in \text{Im} f$ then $y=f(x) \Longrightarrow f \circ f(x)=0$.
So $\text{Ker} f = \text{Im} f$ if $f(0)=0$ and if for all $x$, $f\circ f(x)=0$.
A: If $\phi : V\to W$ is a linear transformation of vector spaces , then $\ker\phi := \{v\in V\mid \phi(v) = 0\in W\}$ and $\textrm{im}\,\phi = \phi(V) := \{w\in W\mid w = \phi(v)\textrm{ for some }v\in V\}$. We see that $\ker\phi\subseteq V$ and $\phi(V)\subseteq W$, so if $V\subsetneq W$ and $W\subsetneq V$, these sets are not even comparable.
If $\phi : V\to V$, then $\ker\phi = \phi(V)$ if $\ker\phi \subseteq \phi(V)$ and $\ker\phi \supseteq \phi(V)$, and I claim that $\ker\phi = \phi(V)\implies\phi(\phi(V)) = \{0\}$. If $\ker\phi \supseteq \phi(V)$, then for all $v\in\textrm{im}\,\phi$, $\phi(v) = 0$. This implies that $\phi(\phi(V)) = \{0\}$.
Moreover, by the rank-nullity theorem, we can say that
$$
\dim V = \dim\ker\phi + \dim\phi(V) = 2\cdot\dim\ker\phi = 2\cdot\dim\phi(V),
$$
so that $\dim V$ is either even or infinite. If $\dim V = n < \infty$, then we can say that $\dim\phi(V) = \dim\ker\phi = n/2$.
