100 dot product riddle
Consider a set of vectors $x_k\in \mathbb{R}^{100}:$
$$ \Big\{ x_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \\0 \end{bmatrix}, \; x_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ \vdots \\ 0 \\0\end{bmatrix}\; x_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 0\\0 \end{bmatrix}, \; \dots,\; x_{99} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots\\1 \\ 0\end{bmatrix},\;x_{100} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 1\\1\end{bmatrix}\Big\} $$
Then $L_1=\big\{(1)_{\times 99},(2)_{\times 98},\dots,(99)_{\times 1}\big\}, \; L_2=\big\{(-1)_{\times 99},(-2)_{\times 98},\dots,(-99)_{\times 1}\big\}$.
Now consider a set of vectors $x_k\in \mathbb{R}^{100}:$
$$ \Big\{ x_1 = \begin{bmatrix} -1 \\ 0 \\ 0 \\ \vdots \\ 0 \\0 \end{bmatrix}, \; x_2 = \begin{bmatrix} 1 \\ -3 \\ 0 \\ \vdots \\ 0 \\0\end{bmatrix}\; x_3 = \begin{bmatrix} 1 \\ 1 \\ -5 \\ \vdots \\ 0\\0 \end{bmatrix}, \; \dots,\; x_{99} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots\\-197 \\ 0\end{bmatrix},\;x_{100} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 1\\1\end{bmatrix}\Big\} $$
Then $L_1=\big\{(-1)_{\times 99},(-2)_{\times 98},\dots,(-99)_{\times 1}\big\},\; L_2=\big\{(1)_{\times 99},(2)_{\times 98},\dots,(99)_{\times 1}\big\}$.
First, let's consider all the possible answers this question could have:
- It it always possible t otell which list is which
- It is always impossible to tell which list is which
- It is sometimes possible and sometimes impossible to tell, depending on the list.
Since both the dimension and amount of vectors can be arbitrary, we can consider 3 1-dimensional vectors resulting in lists [1,1,1] and [-1,-1,-1]. It is clear that [1,1,1] is the correct one, since you can never have 3 real numbers so that the product of any 2 of them is negative.
On the other hand, consider 2 1-dimensional vectors with lists [1,1] and [-1,-1]. Both are possible from vectors (1), (1) and (1), (-1).
Therefore, the answer 3) is correct. Sometimes it is possible to tell, sometimes it isn't. The fun begins when you try to investigate when it is possible to tell...