In a linear space, 0 times an element of the space need not be 0?

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Fractal20
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Homework Statement


Hello, we are starting to get to Banach spaces and thus linear normed spaces in a functional analysis class and I am realizing that I don't have much experience or intuition with these spaces. So I was reading over the requirements for a linear space in my notes and was surprised that there was not a property that 0[itex]\cdot[/itex]x = 0. Is this just implicitly assumed to be true, or is this really not a property of a linear space?

To be more precise, I have an intuitive understanding of what a linear space means if we are considering Euclidean vectors, but if it is just some abstract space that follows the rules of a linear space, then I don't really know what it means to multiply by a scalar. For example, I know what the output of multiplying a vector by a scalar will be but in a more abstract setting it doesn't seem like such a rule needs to be given, only the property that the result will still be in the space. So I am trying to not make the mistake of applying what I know about normal Euclidean vectors to more general concepts.

Homework Equations


The properties of a linear space

The Attempt at a Solution

 
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Yes, if x is any vector then 0x=0 (where the first 0 is the zero scalar and the second is the zero vector). That's a property of ALL vector spaces and a Banach space is a type of vector space. So they may not have felt a need to specifically mention it.
 
So vector space is interchangeable with linear space?
 
Fractal20 said:
So vector space is interchangeable with linear space?

I would think so.
 
The property ##0x = 0## does not need to be listed as part of the definition, as it can be derived from the distributivity property: a vector space must satisfy ##(a+b)x = ax + bx## for all scalars ##a## and ##b## and all vectors ##x##. Choosing ##a=b=0##, this implies that ##(0+0)x = 0x + 0x##. As ##0 + 0 = 0## is true in any field by definition of the additive identity ##0##, the left hand side simplifies to ##0x##, and we have ##0x = 0x + 0x##. Subtracting ##0x## from both sides, we get ##0 = 0x##.