2 Linear Algebra Proofs about Linear Independence

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The discussion focuses on two proofs related to linear independence in linear algebra. The first proof aims to establish that a set of vectors is linearly independent if the equation Ax = 0 has only the trivial solution, but participants express confusion about the necessary steps to connect the definition to the proof. The second proof successfully demonstrates that if a set of vectors is linearly independent, then the last vector cannot be expressed as a linear combination of the preceding vectors, thereby confirming the initial claim. Participants also discuss the relationship between the rank of the matrix formed by the vectors and the dimension of the solution space for homogeneous systems. Clarifications on the definitions and implications of linear independence are sought to strengthen the understanding of the proofs.
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Homework Statement


Proof 1:
Show that S= {v1, v2, ... vp} is a linearly independent set iff Ax = 0 has only the trivial solution, where the columns of A are composed of the vectors in S. Be sure to state the relationship of the vector x to the vectors in S

2. The attempt at a solution
As far as I can tell (from my book and class), the actual definition of linear independence is
"A set of vectors is linearly independent iff there exists only the trivial solution to c1v1 + c2v2+ ... + cpvp = 0, or similarly only a trivial solution exists to the equation Ax = 0 when the columns of A are composed of the vectors in S. "

I know many implications of this property, but not the steps in between the term and the definition. Is there a more basic definition that I need to start this proof out with? Are the two options I gave in my definition of linear independence really different enough to merit in-between steps?

Homework Statement



Proof 2:
Prove or disprove: If {v1, v2, ... vp}, p>=2 is a linearly independent set, then vp is not an element of Span {v1, v2, ... vp-1}. 2. The attempt at a solution

I just took a guess at this one:

Assume that vp}, p>=2 is linearly independent, and let vp be an element of Span {v1, v2, ... vp-1}.

By definition of spanning, vp is a linear combination of {v1, v2, ... vp-1} .

By definition of linear combination, there exists c1v1 + c2v2+ ... + cpvp-1 = vp.

Algebra give us:
c1v1 + c2v2+ ... + cpvp-1 - vp = 0.

This gives a solution other than the trivial solution (where {c1, c2, ... cp} are all zero) to the equation c1v1 + c2v2+ ... + cpvp = 0, because cp must equal -1.

This is a contradiction. {v1, v2, ... vp}, p>=2 can't be a linearly independent set while vp is an element of Span {v1, v2, ... vp-1}. , so if {v1, v2, ... vp}, p>=2 is a linearly independent set, then vp is not an element of Span {v1, v2, ... vp-1}, as desired. Q.E.D.Any help or hints on either of these would be greatly appreciated. The first one I just need a starting point, and the second one I need to know if I made any mistakes, or if that is a valid proof. Thanks a lot.
 
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I agree with the 2nd proof. In the first problem, the statement and the predicate (the definition) seem to be identical -- I am not clear on what is being asked.
 
Regarding proof 1, do you know anything about the dimension of the vector space of solutions to a homogenous system? How does it relate to the rank of the system matrix A? What *is* the rank of your matrix A, since it consists of p linear independent vectors?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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