In my lectures, the relationship $\underline{\hat{e_k}} \cdot \underline{\hat{ e_j}}$ is given to be equal to $\delta_{kj}$ - how is this so?
From my understanding this is equal to $ \left( \begin{array}\\ 0\\ 0\\ 1 \end{array} \right) \cdot \left( \begin{array}\\ 0\\ 1\\ 0 \end{array} \right) $, which would give $\left( \begin{array}\\ 0\\ 0\\ 0 \end{array} \right)$; not the kroenecker delta.
For context, this is discussed whilst going over an introduction to the moment of inertia tensor.
On
The dot products is measuring angles between vectors. Especially, if $a$ and $b$ are unit vectors, then $a\cdot b=\cos\angle(a,b)$, where $\angle(a,b)$ is the angle between $a$ and $b$.
So the angle between the unit vector $e_i$ and $e_i$ is $0^\circ$ (because it's the same vector twice). Note that $\cos(0^\circ)=1$. Therefore
$$e_i\cdot e_i=\cos\angle(e_i,e_i)=\cos0^\circ=1=\delta_{ii}.$$
On the other hand, for $i\not=j$, the unit vectors $e_i$ and $e_j$ are perpendicular to each other (by definition). This means $\angle(e_i,e_j)=90^\circ$ and $\cos 90^\circ=0$. Therefore
$$e_i\cdot e_j=\cos\angle(e_i,e_j)=\cos 90^\circ=0=\delta_{ij}.$$
The dot product is a scalar, not a vector, so you are obvious wrong.
And then you seem to attach specific meanings to $k$ and $j$ when there are just used as variables in that statement.
$\underline{\hat{e_k}}$ (btw: that's an awful lot of indications it's a vector) just means the $k$'th unit vector, with that in mind try to calculate the dot product again.