What is the rank of a linear transformation $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ such that $ \operatorname{Ker} f \leqslant \operatorname{Im}f$? Show an example  for each case with  the matrix associated with a linear transformation!
By the dimension theorem,  $\dim (\operatorname{Ker} f) + \dim (\operatorname{Im} f) = 3$, and $1 \leq \dim (\operatorname{Ker} f) \leq \dim (\operatorname{Im} f)$, 
so $\dim (\operatorname{Ker} f) = 1$, and $\dim (\operatorname{Im} f) = 2$. 
$\dim (\operatorname{Ker} f) + r(f) = 3$
So the rank of the linear transformation is 2. Is my solution correct?
I don't know how to show the examples of the different cases.
 A: Hint You've pointed out that the Rank-Nullity Theorem (what you call the Dimension Theorem) gives $\dim \ker f + \dim \operatorname{im} f = 3$. Both of these quantities are nonnegative integers, so $(\dim \ker f, \dim \operatorname{im} f)$ must be one of $(3, 0), (2, 1), (1, 2), (0, 3)$. Which of these possibilities also satisfies your other condition?
To construct an example in the $\dim \ker f = 1$ case, it is convenient to suppose a transformation satisfying the given conditions exists, work in a basis adapted to our conditions, and build a matrix representation of the transformation in that basis. We can choose:


*

*a (nonzero) vector ${\bf e}_1$ such that $\ker f = \langle {\bf e}_1 \rangle$,

*a vector ${\bf e}_2 \in \operatorname{im} f$ such that $\operatorname{im} f = \langle {\bf e}_1, {\bf e}_2 \rangle$, and

*any vector ${\bf e}_3 \not\in \operatorname{im} f$,


so that $({\bf e}_1, {\bf e}_2, {\bf e}_3)$ is a basis of $\Bbb R^3$.
Now, since $f({\bf e}_1) = {\bf 0}$, the matrix representation of $f$ w.r.t. our basis must have the form
$$[f] = \pmatrix{0&\ast&\ast\\0&\ast&\ast\\0&\ast&\ast\\} .$$
What conditions on $[f]$ does the other requirement, $\operatorname{im} f = \langle {\bf e}_1, {\bf e}_2 \rangle$, impose?
Alternatively, if you have available the Jordan Normal Form, the facts that $\dim \ker f = 1$ and $\ker f \subset \operatorname{im} f$ imply that there is some $\bf v$ such that $f({\bf v}) \in \ker f - \{0\}$. Then, ${\bf v}$ is a generalized eigenvector of $f$ of eigenvalue $0$ (as $f(f({\bf v})) = 0$) but not an eigenvector, so the Jordan normal form of $f$ has a Jordan block of eigenvalue $0$ of size greater than $1$, which leads quickly to the construction of explicit matrices satisfying the given conditions (in fact, all matrices up to similarity).
A: The nullity of a transformation doesn't have to be equal to or greater than 1. E.G - the nullity of the identity transformation is 0 (or generally, the nullity of any injective function is 0).
So you have to take two different cases into account. If you're looking for examples, just look for matrices with rank 2 or 3, it shouldn't be too difficult.
