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I came across a task, which is hard to move for myself (just 1st year of math studies). It is: For what $t\in \mathbb{R}$ set of solutions is a subspace of a $\mathbb{R}^{3}$
$$\\ \\3x_{1}+(1-t^{2}){x_{2}}^{3}-x_{3}=0\\ \\x_{1}-5x_{2}+2(t+1)\left |x_{3} \right |=t^{3}-t$$ I actually solved it for $t=-1$; then the cube term with in first equation vanishes (altogether with absolute value in second) and then the quest is easy. But what for other $t?$ Do you have any ideas? Thanks in advance!

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Clearly the zero vector needs to be part of the solution. This means the latter equation must satisfy $t^3 - t = 0$. This immediately give you all the candidates as $t=0,\ 1,\ -1$.

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    Yes, you are right, but how can I prove that for $t=1, 0$ the set of solutions is not a subspace (I managed to check that for $t=-1$ it is a subspace)?2012-10-30
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    Consider the fact that solutions in a subspace are closed under scalar multiplication. The absolute value in the second line will pose problems.2012-10-30
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    Ok, I understand know (thanks a lot), but one more question. If I want to give a 'writing' proof, would it be enough just to mention (and check) the fact, that absolute value pose problems, or something more is needed?2012-10-30
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    It definitely would not be sufficient for a formal proof (it's more of a sketch). To provide a formal proof, I would explicitly give counter examples. Find a non-zero solution $\mathbf{x}$, and then show that $-\mathbf{x}$ is not a solution.2012-10-30
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    Alright, thank you very much; now there is no problem with it :-)2012-10-30