Does every vector have an additive inverse? I'm taking a linear algebra class at my university and recently we have been talking about the multiplicative inverse of matrices and how some matrices do not have multiplicative inverses (i.e. matrices that aren't square). This made me wonder whether vectors have inverses as well, specifically additive inverses. Isn't the additive inverse of a vector just the negative of that same vector, and isn't the sum of a vector and its additive inverse just the zero vector? In that case, is the statement below true in all cases? And does the zero vector have an inverse?
"Every vector must have an additive inverse, the sum of these two vectors being the zero vector." 
 A: In standard vector spaces you have only addition and scalar multiplication, so the only inverse is the additive inverse.
$$ \mathbf{v}+(-\mathbf{v})=\vec{0} $$
However, in geometric algebra vectors exist as a subset of a larger set of objects including scalars and "multi-vectors" in which a product is defined. This product subsumes the scalar product of standard vector theory.
In this context, some non-zero vectors have multiplicative inverses.
A: Every matrix and every vector have an (additive) inverse. In fact, the set of $n\times n$ matrices for a given $n$ is a vector space, and a vector is defined to be an element of a vector space, which is defined in part by the existence of an additive inverse for every element. 
However, the inverse matrix you're referring to is a multiplicative inverse, which is not guaranteed to exist for every matrix. In general, vector spaces do not have multiplication operators (except scalar multiplication) and so do not have multiplicative inverses or identities. The set of $n\times n$ matrices is a special kind of vector space, called an algebra, which contains a vector multiplication operator (matrix multiplication). The algebra of $n\times n$ matrices is called a unital algebra because of the existence of a multiplicative identity (the identity matrix). However, it is not a division algebra, since there is not a multiplicative inverse for every nonzero element. An example of a division algebra would be the real numbers, since every nonzero real number has a reciprocal, and when the two are multiplied the result is 1.
A: A vector space consists of a set $V$, a field $\Bbb K$, an addition operation $+\colon V \times V \to V$, and a multiplication-by-scalar $\cdot\colon \Bbb K \times V \to V$ satisfying certain properties.
In the particular case of $V={\rm Mat}(n,\Bbb K)$ we have an extra operation of matrix product $${\rm Mat}(n, \Bbb K) \times {\rm Mat}(n,\Bbb K) \to {\rm Mat}(n, \Bbb K)$$which is not in general available in an arbitrary vector space. This is why in general we don't speak of "inverting vectors": one has to be clear about which operation is being considered. In an arbitrary vector space, the only operation guaranteed is the addition $+$. And by definition, every vector $v$ has an "inverse according to $+$": its symmetric $-v$.

There is another algebraic structure a bit stronger than a vector space: an algebra over a field $\Bbb K$ is a vector space equipped with a bilinear product $V \times V \to V$. Our first example of an $\Bbb R$-algebra is ${\rm Mat}(n,\Bbb R)$ with the usual matrix product. In algebras where we have an "unit" element (these are called unital algebras), one can investigate when a given vector has or not an inverse.
A: There are matrices that have no multiplicative inverse but all matrices have an additive inverse. A vector has an additive inverse. The additive inverse of the null vector is the null vector (compare this to your definition). The term inverse is always related to a binary operation.
A: Yes, every vector (even complex ones) has opposite (in your words, additive inverse)
$\vec{u}+(-\vec{u})=0$. Where $\vec{u}=(u_1,\dots,u_n) : u_i \in \mathbb{R}$ (or $\mathbb{K}$ in general)
In fact that is a linear field axiom (take a look).
But every matrix has opposite too.
$A+(-A)=0$ such that $A, -A \in \mathscr{M_{m \times n}}$.
Where $-A$ is defined as follow
$-A_{i,j}=(-1)A_{i,j}$.
Now the inverse (multiplicative inverse) is not defined for vectors but yes for the square matrices whose determinant is not zero.
if $\det{(A)} \neq 0\Leftrightarrow A^{-1}$ exists.
