# Can vectors be inverted?

I wish to enquire if it is possible to solve the below for $C$.

$$B^{-1}(x-\mu) = xc$$

Here obviously $B$ is an invertible matrix and both $c$ and $\mu$ are column vectors. Would the solution be $$x^{-1}B^{-1}(x-\mu) = c$$ is it possible to invert vectors ?

How about if it was the other way

$$B^{-1}(x-\mu) = cx$$

Is there any other way to do this ? Thanks in advance.

• Do you mean for $C$ to be a scalar (or a 1x1 matrix)? That's the only way in which the right hand side product will be defined and of the same size as the left hand side. Oct 6 '12 at 23:16
• @LoganMaingi Yes that is correct. Sorry i probably should have highlighted that. Oct 6 '12 at 23:19
• In general, you should try to denote different kind of objects with different kind of names, as suggestive as possible. So writing the equation as $B^{-1}(v-w)=cv$ would make it easier to understand, at least for me. You also usually write scalars on the left (in commutative context). Oct 6 '12 at 23:22
• @LoganMaingi On another thought i probably could define the problem with CX as well in which case C could really be non 1x1 matrix. For the sake of the argument i would love to find out how that could be solved. Oct 6 '12 at 23:24
• @tomasz Thanks for your suggestion i 'll update my question accordingly. Oct 6 '12 at 23:27

Vectors, in general, can't be inverted under matrix multiplication, as only square matricies can inverses. However, in the situation you've described, it's possible to compute $c$ anyway, assuming the equation is satisfied for some $c$. If we multiply both sides by $X^T$, the result is $x^T B^{-1} (x-\mu) = x^T x c = |x|^2 c$. Assuming x is not the zero vector (in which case any $c$ will still have $xc=0$ so any choice of $c$ should work), we just get $c= \frac{1}{|x|^2} x^T B^{-1} (x-\mu)$.

I must caution that the expression above for $c$ is defined even when there is no solution of the original equation, which will be almost all of the time for randomly generated vectors and matrices. Hence, if you are going to use it, you should check that this works by plugging what you get for $c$ back into the original expression and see if it works. Also, the choice of $x^T \over |x|^2$ is not unique; any row vector $v$ such that $vx=1$ will work equally well in the above expression.

$$B^{-1}(x-\mu) = xc$$ If the matrix $B$ is known, why not write $$(B^{-1} - cI)x = B^{-1}\mu$$ and then worry about whether $B^{-1}-cI$ is invertible?

• I think in the question $B,x,\mu$ are parameters, and $c$ is to be determined. Oct 7 '12 at 0:38

There's no such thing as an inverse of a vector (unless the vector is actually a $1\times 1$ vector, of course).

Otherwise, there would be a solution $C$ for any $B,X,\mu$ (or at least any $X$ "invertible"), but that is obviously not the case (e.g. for any $X$ if we put $B=I$, $\mu$ linearly independent from $X$, there is no $C$).

• What about $\mathbb{C}$ or $\mathbb{H}$? these don't count as vectors? As real vectors complex numbers can be thought of as $2$-vectors, or quaternions as $4$-vectors. Am I wrong? Oct 7 '12 at 0:41
• @JamesS.Cook: Those are algebras over $\mathbb R$. The algebra structure on them are not really all that related to the additive structure. For example $\mathbb C$ is also an algebra over $\mathbb R$ with the multiplication $a*b = \overline{ab}$. Oct 7 '12 at 1:16
• @JamesS.Cook: you're wrong in that the "representation" of complex numbers as 2-vectors and quaternions as 4-vectors has little to do with their respective multiplications (so far as I know...). Complex numbers and quaternions do have natural matrix representations as $2\times 2$ and $4\times 4$ real matrices (with associated natural actions on ${\bf R}^2,{\bf R}^4$, respectively), but not as vectors. Oct 7 '12 at 1:53
• I think I've even heard of complex numbers being defined as the centralizer of $O(2)$ as a subset of $M_{2\times 2}({\bf R})$, or something like that. :) Oct 7 '12 at 2:03
• @tomasz But, in the context of $\mathbb{C}$ or $\mathbb{H}$ multiplication and division of vectors does make sense. Because these are also division rings. Of course these are special cases and yes they do have natural representations as matrices, but that does not negate the fact that we can divide by these particular vectors. Perhaps there is more to your definition of vector than being an $n$-tuple with $n>1$? Oct 7 '12 at 3:19

An inverse rectilinear vector ā' is a vector which is co-directed (in the same direction as) a vector ā and differs from it in magnitude according to: $$|\bar{a'}|=\dfrac{1}{|\bar{a}|}$$ Projections on the coordinate axes of inverse rectilinear vectors are equal according to: $$\quad a'_{x}=\dfrac{a_{x}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}};\quad a'_{y}=\dfrac{a_{y}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}\quad a'_{z}=\dfrac{a_{z}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}$$ An example of solving problems with this vector can be found here https://en.wikipedia.org/wiki/Talk:Cross_product#Cross_product_does_not_exist https://doi.org/10.5539/jmr.v9n5p71
If we have: $$V= [v_1, v_2 ... v_n],$$ then:
$$V^{-1} = [v_1^\dagger, v_2^\dagger, ... v_n^\dagger].$$