Proving an object is a $4$-vector given its inner product with a $4$-vector is a scalar

Question

Theorem: Suppose $A_{\mu}$ is a $4$-vector and $B^{\mu}$ is an object with $4$ components. If $A_{\mu}B^{\mu}$ is a scalar then $B^{\mu}$ is a $4$-vector.

I have been stuck on trying to prove this theorem for quite a while (See Ref.1) and haven't made much progress. I think this is due to my lack of understanding on what it means to be a $4$-vector.

What I have tried so far is to use the invariant quantities $A_{\mu}A^{\mu}$ and $A_{\mu}B^{\mu}$ to show $B_{\mu}B^{\mu}$ is an invariant quantity.

I would like to know if this approach makes sense, since the only quantifiable property of a $4$-vector I can think of is its Minkowski inner product with itself being invariant.

I am new to this subject and would be really grateful for any help or insights.

References:

1. L. Susskind & A. Friedman, Special Relativity and Classical Field Theory: The Theoretical Minimum , 2017; p. 181 section 5.5 .

Answer 1

From $^{\prime\prime}$Tensor Calculus$^{\prime\prime}$ by J.L.Synge-A.Schild, Dover Edition 1978 :

1.6. Tests for tensor character. The direct test for the tensor character of a set of quantities is this: see whether the components obey the law of tensor transformation when the coordinates are changed. However, it is sometimes much more convenient to proceed indirectly as follows.

Suppose that $\,A_{r}\,$ is a set of quantities which we wish to test for tensor character. Let $\,X^{r}\,$ be the components of an arbitrary contravariant tensor of the first order. We shall now prove that if the inner product $\,A_{r}X^{r}\,$ is an invariant, then $\,A_{r}\,$ are the components of a covariant tensor of the first order. We have, by the given invariance,

\begin{equation} A_{r}X^{r}\boldsymbol{=}A^{\prime}_{r}X^{\prime\, r} \tag{1.601.}\label{1.601.} \end{equation} and, by the law of tensor transformation, \begin{equation} X^{\prime\, r}\boldsymbol{=}X^{s}\dfrac{\partial x^{\prime\, r}}{\partial x^{s}} \tag{1.602.}\label{1.602.} \end{equation} Substituting this in the right-hand side of \eqref{1.601.}, rearranging, and making a simple change in notation, we have \begin{equation} \left(A_{s}\boldsymbol{-}A^{\prime}_{r}\dfrac{\partial x^{\prime\, r}}{\partial x^{s}}\right)X^{s}\boldsymbol{=}0 \tag{1.603.}\label{1.603.} \end{equation} Since the quantities $\,X^{s}\,$ are arbitrary, the quantity inside the parentheses vanishes; this establishes the tensor character of $\,A_{r}\,$, by \eqref{1.402.}

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

A set of quantities $\,T_{r}\,$ are said to be the components of a covariant vector if they transform according to the equation \begin{equation} T_{r}\boldsymbol{=}T_{s}\dfrac{\partial x^{s}}{\partial x^{\prime\, r}} \tag{1.402.}\label{1.402.} \end{equation}