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I want to solve an equation in form of

$$a^Tx = \alpha \ \ \ \ \ \ \ \ \ \ (1)$$

where $a, x \in \mathbb R^n$ and $\alpha \in \mathbb R$.

If the equation was in form of

$$A^Tx = a$$

where $A \in \mathbb R^{n\times n}$ and $a \in \mathbb R^n$

I know I can solve it by pre-multiplying $A^{-T}$ on both sides. But for equation (1), since I cannot inverse the vector $a$ I have no clue how to solve it.

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Let $\mathbf{x}=(x_1,\ldots,x_n)^\top\in\Bbb{R}^n$, $\mathbf{a}=(a_1,\ldots,a_n)^\top\in\Bbb{R}^n$, and $b\in\Bbb{R}$. Then $$ \mathbf{a}^\top\mathbf{x} = b \implies \sum_{i=1}^{n} a_ix_i = b. $$ This is a linear equation with $n$ variables, i.e., the $x_i$'s. You cannot find a solution of $\mathbf{x}\in\Bbb{R}^n$.

In contrast, $A\mathbf{x}=\mathbf{b}$, is a system of $n$ linear equations with $n$ variables. In this case, under certain circumstances you can find a unique, none, or infinite solution(s) for $\mathbf{x}$.

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Generally speaking this equation is geometrically seen as a n-dimensional hyperplane in $R^n$ which is perpendicular to $a^T$.So you can't really solve for $x$.Instead you may find a parametric description for each $x_i$.

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One thing you can do is to solve the normal equations instead. You can do this by multiplying both sides with transpose of the matrix $a^T$ (from the left) :

$$\underset{\text{a matrix}}{\underbrace {aa^T}}x = a \alpha$$

both $x$ and $a\alpha$ are vectors so you are probably more confident trying to solve this.

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