Can I use eigenvalues to find the inverse of a vector?

Question

I have two 1D matrices (say dimension 1xn) called A and B. Multiplying these: A . B = M. Where M is a scalar. Knowing B and M, can I find A?

One cannot take the inverse of a vector, but is it possible to use eigenvalues? as the form for eigenvalues is u=(lambda)v, where u and v are vectors!

Lookinf forward to any helpful answer!

Kind regards, M

Answer 1

You would not be able to uniquely identify $A$. We can see this already by considering two and three dimensions.

$$A\cdot B = \|A\| \|B\| \cos(\theta)$$ where $\theta$ is the angle measured between $A$ and $B$. Note that in two dimensions, if the angle is swapped with $-\theta$ then we have the same result. In three dimensions there are more degrees of freedom.

For example, suppose $A=(1,0)$ and $B=(0,1)$. $A\cdot B =0$ but this is also true for $A'=(-1,0)$ or $A'=(20000,0)$. The dot product can reduce the number of allowable solutions, but unless it is a simple one dimensional system, you cannot determine $A$ from it.

Answer 2

Suppose $A=\begin{pmatrix}b_1\\b_2\\\vdots\\b_n\end{pmatrix}$ and let $B =\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}$. Then $$ A^\top B = M \quad\text{or}\quad a_1x_1+a_2x_2+\ldots+a_nx_n=M. $$

1. If $a_i=0$ for every $i \in \{1,2,\ldots,n\}$ and $M=0$ then $B$ is an arbitrary vector.

2. If $a_i=0$ for every $i \in \{1,2,\ldots,n\}$ and $M \neq 0$ then the is no $B$ such that the required equality holds.

3. Without loss of generality assume that $a_1 \neq 0$, then $x_1 = \frac{1}{a_1}(M-a_2x_2-\ldots-a_nx_n)$. Here $x_2,\ldots,x_n$ are so called free variables. So, $$ B = \left(\frac{1}{a_1}(M-a_2x_2-\ldots-a_nx_n),x_2,\ldots,x_n\right), \quad\text{ where $x_2,\ldots,x_n \in \mathbb{R}$}. $$