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Let's consider $A$ the subspace of $\mathbb R^4$ given by $A=\mathrm{Span}(Y_1,Y_2)$ where

$$\begin{cases} Y_1=(0,1,\sqrt 2,\sqrt 3), \\ Y_2=(1,0,\sqrt 3,\sqrt 2).\end{cases}.$$

I was wondering about the answer of the following question:

Does there exist a rational subspace $B$ of $\mathbb R^4$ of dimension $2$, such that $\dim(A\cap B)\geqslant 1$?

Some context.

  • By rational subspace, I mean a subspace of $\mathbb R^4$ which admits a rational basis, in other words a basis $(a,b)$ with $a,b\in\mathbb Q^4$.

  • I conjecture the answer to be no, but I can't prove it so far.

  • We can reformulate the problem the following way: does there exist a vector $v\in A\setminus\{0\}$ such that

$$\exists \xi_1,\xi_2\in\mathbb R,\quad \exists r_1,r_2\in\mathbb Q,\quad v=r_1\xi_1+r_2\xi_2\quad ?$$

  • We can write the problem in coordinates, which leads us to find $\alpha,\beta,\xi_1,\xi_2\in\mathbb R$ and $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2\in\mathbb Q$, such that

$$\alpha Y_1+\beta Y_2=a \xi_1+b\xi_2,$$

which gives

$$\begin{cases} \beta=a_1\xi_1+b_1\xi_2 \\ \alpha=a_2\xi_1+b_2\xi_2 \\ \alpha\sqrt 2+\beta\sqrt 3=a_3\xi_1+b_3\xi_2 \\ \alpha\sqrt 3+\beta\sqrt 2 = a_4\xi_2+b_4\xi_2,\end{cases}$$

but it does not seem to help at all.


Since I am quite stuck with this problem, any help, hints, or new ways to look at the problem would be much appreciated.

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    $\begingroup$ @AlexRavsky Absolutely, thanks. I have edited to correct the typo. $\endgroup$ – E. Joseph Jan 13 at 9:22
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$$ (\sqrt2 + \sqrt3) (Y_1-Y_2) = \pmatrix{ 1\\ -1 \\ \sqrt2 + \sqrt3 \\ -(\sqrt2 + \sqrt3)} = \pmatrix{ 1\\ -1 \\0\\0} + (\sqrt2 + \sqrt3)\pmatrix{0\\0\\1\\-1} $$

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    $\begingroup$ $Y_1-Y_2=(-1,1,\sqrt2-\sqrt3,\sqrt3-\sqrt2)=(-1,1,0,0)+(\sqrt2-\sqrt3)(0,0,1,-1)$. $\endgroup$ – Gerry Myerson Oct 24 '18 at 21:40

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