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In a vector space, how much null vectors exist?
\(\text{The most common use of the term null vector refers to the vector consisting of all 0's.}\\ \text{Thus in a given vector space there exists only one of these.}\\ \\ \text{A less common use refers to a vector }x \text{, such that for a given matrix }A, A x = 0\\ \\ \text{If there is such a vector other than the vector of all 0's,}\\ \text{ then there are infinity of them as any scalar multiple of }x\\ \\ \text{will also satisfy the definition.}\)
In any given vector space there is only one null vector.In V3r it is (0,0,0).- normally written as a column vector.Think of it as the equivalent of the 'zero' of the real number system. It is required because clearly,for example, if you subtract a vector from one with its equal and opposite magnitude ,you get the vector equivalent of zero,which needs to be meaningfully represented and to have the same properties as the other vectors in the space.Just as you need zero for numerical and algebraic operations,you need the null vector to be able to carry out a full range of vector operations.
You can easily check that the null vector obeys the requirement of a vector space,which is simply that k(a+b)= ka+kb where k is a constant and a,b are any 2 vectors in the given vector space.(You can also easily check that it obeys the dot product.)
A quick proof for the vector space---
Let a,b be 2 linearly independent vectors of a given space.Let the null vector be denoted by 0.
We have k(a+b)=ka+kb.
Then k(a+0) = ka +k0 = ka,since a+0 =a and k0=0
Similarly k(b+0) =kb+k0 =kb,since b+0=b.
and k(a+b+0) =ka+kb+k0=ka+kb,since a+b+0= a+b.
A proof for the dot product is also easy,have a try.
