Inverse of a vector
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 $v$?
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$\begingroup$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 $2$.
$\endgroup$ $\begingroup$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.
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