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In $\mathbb R^3$, consider the following statements about the subset $$E = \{(1,0,0),(0,1,0),(0,0,1),(1,1,1),(1,1,0)\}$$

Which of the following is/are correct:

  1. $E$ is linearly dependent.
  2. Any three vectors in $E$ are linearly independant.
  3. Any four vectors in $E$ are linearly dependent.

My attempt : (1,1,1) = (1,0,0)+(0,1,0)+(0,0,1), therefore, 1. is true.

Also, since $\dim \mathbb R^3 = 3$, any four vectors in $E$ would be linearly dependent, therefore 3. is true.

I am stuck on 2., as it asks about any 3 elements of $E$. And there are 10 choices for 3 elements out of 5. The only method i can think of is to reduce all these triplets in a matrix, and if the rank is 3 then they are linearly independent, but that will take much time.

How to reduce cases? Or to solve in one matrix altogether, if possible.

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    Please use $\LaTeX$ – 2017-01-26
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    Sorry, but i am writing this from my phone, and it is hanging if i try with $. – 2017-01-26
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    @Aweygan : What is used here is MathJax, not LaTeX. LaTeX has zillions of features not found in MathJax. Whoever masters MathJax and thinks they know LaTeX will suffer a severe shock when they encounter actual LaTeX and find they don't know it. – 2017-01-26

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For $2)$, observe that $$\begin{pmatrix}1\\1\\1 \end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1 \end{pmatrix}$$

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    What if there were 50 vectors in E, and we were asked if any 25 of them would be linearly independent? All i am saying is, other than keen observation, is there any other method? – 2017-01-26
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    @Shobhit Well if the dimension of the vector space is $\leq 24$, then the answer is simple. If the dimension is $\geq 25$, then yes you would have to determine linear independence via checking each $25$-element subset, or getting lucky and finding one which isn't. – 2017-01-26
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    Ok, understood. Thank u. – 2017-01-26