Orthogonal vectors and linear systems

Question

Let us suppose we want to solve, with respect to x, the following equation

$\mathbf{a}^\intercal\mathbf{b}\;x = 0$

where $\mathbf{a}, \mathbf{b} \in \mathbb{R}^{n} \setminus \{ 0 \}$. It seems clear to me that the solution is $x \in \mathbb{R}$ if the two vectors are orthogonal, $x = 0$ otherwise. Let us now consider a vector $\mathbf{c} \in \mathbb{R}^{n}$ such that $\mathbf{c}\mathbf{a}^\intercal = \mathbb{I}_n$. If we multiply both sides of the equations by $\mathbf{c}$, we get

$\mathbf{b}\;x = 0$

that seems to have just the solution $x = 0$, even when the two vectors $\mathbf{a}$ and $\mathbf{b}$ are orthogonal. For sure, I have committed an error somewhere, doing something I was not allowed to do, but I really cannot understand what. Does someone spot the error? Thanks!

EDIT: I have edited the question to fix two errors:

• $\mathbf{c} \in \mathbb{R}^{n}$ instead of $\mathbf{c} \in \mathbb{R}$
• $\mathbf{c}\mathbf{a}^\intercal = \mathbb{I}$ instead of $\mathbf{c}\mathbf{a}^\intercal = 1$

Answer 1

Suppose that there exists $c\in\mathbb{R}^n$ such that $ca^T=\mathbb{I}$. Since $0=a^Tb$ we have $$ (0, 0,\ldots, 0)^T=ca^Tb=\mathbb{I\;} b=b, $$ and so $a^Tbx=0$ implies $0\times x=0$ and so $x\in \mathbb{R}$.

Answer 2

For $\mathbf{ca}^T=\mathbb{I}$ you need $$c_ia_j=0\quad\textrm{for}\quad i\neq j\\\textrm{and}\quad c_ia_i=1$$ These requirements are not compatible.

Answer 3

If c is a scalar then

$ca^T=(ca_1, ca_2,\ldots, ca_n)^T $ and unless $a_1 = a_2 = ... = a_n \not = 0 $

$ca^T\ne 1. $