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The set {0} with the + and × operations defined as 0+0=0 and 0×0=0 sarisfies the properties to be a field, so a vector space can be constructed over it, with the property $0x = x $ where x is a vector.

In my book it states that the vector $x $ by itself forms a linearly dependent system iff $x=0$. This seems to lead to a contradiction: if we assume that the vector space is over {0} with the properties said before, then $x =0$ is also linearly independent since the only way to get the null vector from it is multiply it by 0...

So does this mean that it is linearly dependent and independent at same time? Which would mean that lin. dependence is not the negation of lin. independence... Or is the book wrong? Or am I making a mistake?


marked as duplicate by Jyrki Lahtonen, Arthur, hardmath, Community Aug 18 '18 at 13:05

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    $\begingroup$ In the definition of a field the set of non-zero elements of the field forms a group. A group cannot be empty. Therefore $\{0\}$ is not a field in spite of your argument. The rest of your question is moot. $\endgroup$ – Jyrki Lahtonen Aug 18 '18 at 12:33
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    $\begingroup$ There is no "null" field, since a field must contain $0$ and $1$, with $0\ne 1$, $\endgroup$ – quasi Aug 18 '18 at 12:33
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    $\begingroup$ $\{0\}$ is not a field, it's a ring. In a field $0$ and $1$ can't be the same element. $\endgroup$ – Mark Aug 18 '18 at 12:34
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    $\begingroup$ @quasi Its a portuguese book, Algebra Linear: Espaços Vectoriais, A. Monteiro. $\endgroup$ – Carla Kis Aug 18 '18 at 12:59
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    $\begingroup$ @Carla Kis: If the author allows $\{0\}$ to be considered a field (which I've never seen in other standard Linear Algebra textbooks), it's not a field of any importance in Linear Algebra, so don't fret about issues that are unique to that special case. $\endgroup$ – quasi Aug 18 '18 at 13:04