Can More Vectors Than Dimensions Be Linearly Independent?

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The discussion centers on proving that if a set S contains m vectors in n dimensions and is linearly independent, then the number of dimensions n must be greater than or equal to m. Participants explore the contradiction method by assuming m is greater than n, leading to the conclusion that if there are more vectors than dimensions, at least one vector can be expressed as a linear combination of others, which contradicts linear independence. Clarifications are made regarding the definitions of linear combinations and the implications of having more equations than unknowns in the context of vector spaces. The conversation emphasizes the importance of understanding the relationships between vectors and their independence in linear algebra. The thread highlights the complexities of proving linear independence in higher-dimensional spaces.
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Homework Statement


prove, if S has m vectors in n dimensions and S is linearly independent, then n>=m

Homework Equations





The Attempt at a Solution


so far I've come up with:

there is no combination of vectors in S such that their sum is the zero vector,

there exists a vector which cannot be expressed in terms of a linear combination of other vectors

so I've started to assume that m>n (contradiction method)
however I am stuck here.

im thinking of saying that if m>n, then there are more equations than unknowns in the system, but i don't know if that is helpful, and I am stuck.
 
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choob said:

Homework Statement


prove, if S has m vectors in n dimensions and S is linearly independent, then n>=m

Homework Equations





The Attempt at a Solution


so far I've come up with:

there is no combination of vectors in S such that their sum is the zero vector,
I hope you mean "except the trivial combination with all coefficients equal to 0"!

there exists a vector which cannot be expressed in terms of a linear combination of other vectors
It's not clear to me what this means. What vectors are you talking about? Any vector can be expressed as a linear combination of some other vectors!

so I've started to assume that m>n (contradiction method)
however I am stuck here.

im thinking of saying that if m>n, then there are more equations than unknowns in the system, but i don't know if that is helpful, and I am stuck.
I have no idea what "equations" you are talking about. The problem said there were m vectors, not m equations.

Since the set of m vectors is linearly independent, any subset is also linearly indepependent. In particular, if m> n, then any subset of n vectors is linearly independent and so a basis. Now choose a vector in the set, v, that is NOT in that subset (which you can do since m> n). It can be written as a linear combination of the n basis vectors and you can use that to show that that subset of n+1 vectors, v together with the original n vector subset, that is NOT independent, a contradiction.
 
the equations correspond to the matrix which has to be reduced to give the coefficients for each vector, such that their sum is zero, if the system is linearly independent

other than that, ill post again later when i can wrap my head around what you said at the end haha

edit:

what i meant in that there exists a vector which cannot be expressed in terms of the other vectors in that set, is that if there exists v1+v2+..+v3=v4, where all those v's are vectors, then the system would be linearly dependent, because v4-(v1+v2+..+v3)=0
 
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