Finding a linear transformation $T$ such that $T(v) \cap n(T) \neq \lbrace 0 \rbrace$ My professor gave us this problem as a final review, and I am confused what it is asking. If someone could put into words what the question is, that would be helpful. 

Give an example of a linear transformation $T: V \rightarrow V$  such that $T(V) \cap n(T) \neq \lbrace 0 \rbrace$, where $n(T)$ means the nullspace. 

We solved this using matrices... And it seem to me, that it is asking if there is a linear transformation for which the elements in the image, and the null space share an element in common. Would any element of the nullspace be a member of both? 
 A: You're thinking the right things.  Here's how I conceptualize it.
Suppose that you have a vector in $v \in T(V) \cap n(T)$.  Since $v \in n(T)$, you know that $T(v) = 0$ and since $v \in T(V)$, you know that there is some $u \in V$ (not necessarily unique) such that $T(u) = v$.  Like this:
$$
u \overset{T}{\longmapsto} v \overset{T}{\longmapsto} 0
$$
So you have to think about a linear transformation $T$ that eventually sends a vector to zero (after two applications of $T$ but not after one).
The simplest example that I can think of:

 In the standard basis for $\Bbb{R}^2$, $u = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $v = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, and $[T] = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$.

A: An example for this is the following map.
Define $T: V \to V $ such that for $v$ in the basis of $V$
$T(v) =\begin{cases} 0 & if \text{ i is even} \\ v_{i-1} & if \text{ i is odd}
\end{cases} $
the null pace of $T$ is the set of all $v_i$ for $i$ even and $T(V)$ is also the same set.
$\implies T(V) \cap n(T) = $set of $v_i$ for $i$ even.
A: Your interpretation is correct.
Re your question, in general, elements of the null space are not in the image. For example, for any vector space $V$, every element is in the null space of the zero map, but only $0$ is the image.
We can produce examples of transformations with nontrivial intersection deductively, and in fact an easy line of reasoning gives us a characterization of such transformations. 
Let $V$ be a finite-dimensional vector space and $T$ a linear transformation $V \ to V$. Since the intersection $T(V) \cap n(T)$ is nontrivial, so is $n(T)$, and hence $T$ has at least one zero eigenvalue. Now, consider the Jordan Canonical Form $[T]$ of $T$:
Case 1 If all of the Jordan blocks of $[T]$ of eigenvalue $0$ have size $1 \times 1$, say there are $k$ of them, then (after possibly reordering the blocks) $[T]$ has the form $A \oplus 0_{k \times k}$, where $A$ is an invertible $(n - k) \times (n - k)$ matrix. (In order to talk about the Jordan Canonical Form this way, we must be willing to work in some extension of the underlying field $\mathbb{F}$ of $V$ if necessary, but this poses no problem here, as it doesn't affect our conclusions.) With respect to the given basis $(E_a)$, we have $T(V) = \langle E_1, \ldots, E_{n - k} \rangle$ and $n(T) = \langle E_{n - k + 1}, \ldots, E_n \rangle$, and so the intersection is trivial.
Case 2 Suppose $[T]$ has a Jordan block $J$ of eigenvalue $0$ of larger size, say, $m \times m$. Then, as a map $\mathbb{F}^m \to \mathbb{F}^m$, in the standard basis $(E_1, \ldots, E_m)$ we have $J(\mathbb{F}^m) = \langle E_1, \ldots, E_{m - 1} \rangle$ and $n(J) = \langle E_1 \rangle$, so the intersection is $\langle E_1 \rangle$. Passing back to $[T]$ itself, by virtue of how the Jordan Canonical Form behaves (namely, it is a decomposition into a direct sum of linear transformations), the intersection for $J$ is a subset of the intersection for $T$. In particular, each such Jordan block, that is each matrix
$$\begin{pmatrix}0 & 1 &   & & & \\
                   & 0 & 1 & & & \\
                   &   & \ddots & \ddots & & \\
                   &   &        &      \ddots & 1 & \\ 
                   &   &        &        & 0 & 1 \\
                   &   &        &        &   & 0
\end{pmatrix},$$ regarded as a linear transformation of $\mathbb{F}^m$, is itself an example of a linear transformation for which the intersection is nontrivial.
We can conclude the following:
Proposition Let $V$ be a finite-dimensional vector space and $T$ a linear transformation $V \to V$. Then, the intersection $T(V) \cap n(T)$ is nontrivial iff the Jordan Canonical Form $[T]$ of $T$ contains a block of eigenvalue $0$ and size at least $2$. Moreover, the number of such blocks in $[T]$ is $\dim (T(V) \cap n(T))$.
Remark By the Rank-Nullity Theorem, the dimension of the intersection must satisfy
$$\dim (T(V) \cap n(T)) \leq \left\lfloor \frac{\dim V}{2} \right\rfloor.$$
