Inverse of a vector

Question

How would we calculate the inverse of a $3$-vector $v$ like we would do with the matrix? Basically the inverse of $1\times3$ (?) matrix.

Would we get the negated value of the vector $-v$ or $1/v$? If we would get $-v$, what is correct term for $1/v$ in relation to $v$?

Answer 1

As @EeveeTrainer states, it makes sense to talk about additive inverses. Consider you're using a 1x3 vector to model displacement in $\mathbb{R}^3$. Specifically, suppose you're walking down a straight road and you move 3km east, disregarding sea-level elevation (assuming the road is perfectly flat). One might model this as the vector $\vec{v} = (3,0,0).$

Now consider you want to do the opposite. One interpretation could be that the inverse of this 1x3 vector is the 1x3 vector $-\vec{v} = (-3,0,0)$.

It is important to note that the sum of these two vectors is 0, the additive identity.

Answer 2

See Moore-Penrose inverse (for a vector). In particular, if $v \neq 0$ and $w=\lVert v \rVert^{-2} v $ then $\langle w, v \rangle = w^{\mathrm t} v = 1$ (i.e. it is a “left” inverse) and $w$ minimizes the Frobenius norm $\lVert v w^{\mathrm t} - 1 \rVert^2$ among all vectors (i.e. it is the best “right” inverse). In dimension $3$ this minimal square Frobenius norm is always $2$.

Answer 3

It all depends on what is meant in the question, I will propose a possible geometric interpretation which could be the following: given a certain vector $\vec{a}$, we intend to find the set of vectors $\vec{x}$ such that given

$$\vec{a} \cdot \vec{x} = 1$$

, then we can easily find that this set is given by the following decomposition along:

$$x = \frac{\vec{a}}{|a|^2}+\vec{v}$$

where $\vec{v}$ is a generic vector orthogonal to $\vec{a}$. For example, in $R^2$ we can define the set of vectors $\frac{1}{\vec{a}}$:

$$\frac{1}{\vec{a}} = \vec{x}(u) = \frac{\vec{a}}{|a|^2}+u\left[\begin{array}{cc}0&-1\\1&0\end{array}\right] \vec{a}$$ where u is a parameter.

That set describes an orthogonal line to $\vec{a}$ distant $1/|\vec{a}|$ from the origin.

As another example, in $R^3$ we would have (now $\vec{u}$ is a generic vector which operates as parameter):

$$\vec{x}(\vec{u})= \frac{\vec{a}}{|a|^2} + \vec{u} \wedge \vec{a}$$

that describes an orthogonal plane to $\vec{a}$ and distant $1/|a|$ from the origin. In general we could think that the operation $1/\vec{a}$ is a compact way to represent the hyperplane orthogonal to $\vec{a}$, distant from the origin $1/|\vec{a}|$ and in particular passing through the point $\vec{a}/|\vec{a}|^2 $.

PS: This concept seems similar to that discussed in What is vector division?. Indeed, could we think to decompose the division of two vectors as the inversion of the divisor vector [which produces a matrix] followed by the multiplication with the dividend vector?