Component-free formula for the determinant of a tensor Consider a unit vector $\mathbf{a}\in\mathbb{R}^3$ and the associated second-order tensor $\mathbb{A}=\mathbf{a}\otimes\mathbf{a}$. Is there a component-free formula for the determinant of this tensor? For instance, the trace operator is such that $\mathrm{trace}\,\mathbb{A}=\mathbf{a}\cdot\mathbf{a}$ where $\cdot$ denotes the usual inner product between two vectors.
More generally, consider an orthonormal basis of $\mathbb{R}^3$, $\mathbf{m},\mathbf{n},\mathbf{p}$, what is the determinant of $\mathbb{A}=\mathbf{m}\otimes\mathbf{n}+\mathbf{n}\otimes\mathbf{m}+\mathbf{p} \otimes \mathbf{p}$?
 A: You may find the perspective on linear maps from geometric algebra and calculus interesting.
Geometric algebra is a clifford algebra based on some real vector space.  There is a grading structure involved, and a geometric product of vectors so that if $a, b$ are vectors, $ab \equiv a \cdot b + a \wedge b$ is a meaningful statement: this is a "multivector" with a grade-0 part (the dot product) and a grade-2 part (the wedge product).
Linear operators are typically expressed as functions of vectors or $k$-vectors.  For instance, your map $\mathbb A$ would be written as
$$\mathbb A(b) = (a \cdot b) a$$
for any vector $b$.
Traces are taken using differentiation with respect to the vector argument.  Just as $\nabla$ is often taken to mean "vector derivative with respect to position vector," take $\nabla_b$ to mean "vector derivative with respect to given vector $b$".  Thus, the trace is just a divergence of a linear operator:
$$\mathrm{Tr } \ \mathbb A = \nabla_b \cdot \mathbb A(b) = a \cdot a = a^2$$
There is an associated bivector invariant
$$\nabla_b \wedge \mathbb A(b) = a \wedge a = 0$$
but it's trivial here because $\mathbb A$ is symmetric.
Determinants are taken using a concept dubbed "outermorphism".  A linear operator can be extended to act on $k$-vectors across wedge products like so:
$$\mathbb A(b \wedge c) \equiv \mathbb A(b) \wedge \mathbb A(c)$$
The determinant in 3d is defined using some 3-vector quantity $i$, a pseudoscalar.  $i$ can be defined as any 3-vector $i = b \wedge c \wedge d$ for 3 linearly independent vectors $b, c, d$.  $i$ can be seen as an oriented volume.
The determinant is then defined as the scalar $\alpha$ such that
$$\mathbb A(i) = \alpha i \implies \mathbb A(b \wedge c \wedge d) = \alpha (b \wedge c \wedge d)$$
This captures the geometric meaning of how determinants scale or reorient the volume elements of a given space.
Of course, we can see right away that this is zero for your original map $\mathbb A(b) = (a \cdot b)a$ because $a \wedge a = 0$.
For your second map,
$$\mathbb A(b) = (m \cdot b) n + (n \cdot b) m + (p \cdot b) p$$
I conveniently choose $i = m n p $ (using the geometric product, this is equivalent to wedges for orthogonal vectors) and we get
$$\mathbb A(i) = \mathbb A(m) \wedge \mathbb A(n) \wedge \mathbb A(p) =n \wedge m \wedge p = -i$$
Therefore, the determinant is $-1$.

I hope this has convinced you that a geometric algebra (or at least, an exterior algebra) view on linear maps is useful in practice.  GA is often said to be a quotient of the tensor algebra, with the implication that it is less useful.  I disagree vehemently.  GA is the algebra of useful geometric primitives.  The tensor algebra is the wider algebra of linear maps.  Both are useful.  And while objects in GA always have associated maps, objects in the tensor algebra don't necessarily have geometrically meaningful counterparts.  Keeping these separate maintains a great deal of clarity into what you're actually dealing with.
