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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.

Comic Book Guy
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  • 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. – Logan M Oct 06 '12 at 23:16
  • @LoganMaingi Yes that is correct. Sorry i probably should have highlighted that. – Comic Book Guy Oct 06 '12 at 23:19
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    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). – tomasz Oct 06 '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. – Comic Book Guy Oct 06 '12 at 23:24
  • @tomasz Thanks for your suggestion i 'll update my question accordingly. – Comic Book Guy Oct 06 '12 at 23:27

6 Answers6

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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.

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

Michael Hardy
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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 $C$).

tomasz
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    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? – James S. Cook Oct 07 '12 at 00:41
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    @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 $a*b = \overline{ab}$. – kahen Oct 07 '12 at 01:16
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    @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. – tomasz Oct 07 '12 at 01:53
  • 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. :) – tomasz Oct 07 '12 at 02:03
  • @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 $n>1$? – James S. Cook Oct 07 '12 at 03:19
  • @JamesS.Cook: It negates it in the sense that the quaternion multiplication as you can define it for vectors has nothing to do with matrix multiplication. Vectors (in matrix calculus context, anyway), are usually interpreted as $1\times n$ (or $n\times 1$) matrices and have associated algebraic properties, including matrix multiplication. The equation in the question is really about matrix calculus, with matrix multiplication, nothing else, and that's the sense in which OP was looking for an inverse. – tomasz Oct 07 '12 at 11:27
  • @JamesS.Cook: Continuing, you could just put on the vectors the structure of ${\bf R}^n$ (ring product, coordinate multiplication and addition), and then a lot of elements would be invertible in sense of that multiplication; in fact, I'm pretty sure you can put some multiplicative structure on any set which would admit inverses, but that is beside the point. – tomasz Oct 07 '12 at 11:30
  • @tomasz fair enough, I was reacting to the first sentence in your post without paying enough mind to the second. I think we all agree that in the context of an unital algebra certain vectors are multiplicative units. Moreover, if we identify vectors with their representative matrices then the inverse may well be the usual inverse of the matrix representation. But, that is not the context of the original question here so I'll shut up with that. – James S. Cook Oct 07 '12 at 18:19
  • In the geometric algebra, a vector has a multiplicative inverse. – NicNic8 Nov 07 '21 at 16:42
  • @NicNic8: Any time you have an algebra over a field, it is also a vector space over that field. But having multiplication (or multiplicative inverses) is a feature of the algebra, not the vector space. – tomasz Nov 07 '21 at 21:39
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Perhaps this definition of the inverse vector will help you:

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

Yujin -X
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In a quantum mechanics book, I've seen it defined this way:

If we have: $V= [v_1, v_2 ... v_n],$ then:

$$V^{-1} = [v_1^\dagger, v_2^\dagger, ... v_n^\dagger].$$

amWhy
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Francky_V
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I think there is also a very simple answer which is that multiplying a vector of $1\times n$ by another of $n \times 1$ gives you a number, and $n\times 1$ by $1\times n$ gives you an $n\times n$ matrix.

That is, insofar as we look at standard multiplication (alike the matrix ones), the answer is no.

Mah Neh
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