On proving Linearly independence

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SUMMARY

This discussion focuses on proving the linear independence of vector sets in a vector space. It establishes that if vectors {a, b} are linearly independent and vector c does not belong to the span of {a, b}, then the set {a, b, c} is also linearly independent. The proof involves demonstrating that the only solution to the equation c1a + c2b + c3c = 0 is c1 = c2 = c3 = 0. Additionally, it discusses how to show that a linear combination w of vectors in set T, which are linear combinations of vectors in set S, can be expressed as a linear combination of the vectors in S.

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franz32
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Hello. I want to ask questions... I hope you can guide me
in showing the proof.

1. Let a, b and c be vectors in a vector space such that {a, b} is linearly independent. Show that if c does not belong to span {a, b}
, then {a, b, c} is linearly independent.

I know that is {a,b} is l. independent, it implies that
c1a + c2b = 0. That is, c1 = c2 = 0.
What does it mean (imply) when c is not in span {a,b}?
How will I show the essence of the proof?

2. Let S = {u1, u2, ..., uk) be a set of vectors in a vector space, and let T = {v1, v2, ..., vm}, where each vi, i = 1, 2, ..., m, is a linear combination of the vectors in S. Show that

w = b1v1 + b2v2 + ... + bmvm
is a linear combination of the vectors in S.

How will I show the essence of the proof? I don't understand the meaning (implication)of the first sentence.
 
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For question 1, the statement implies that c is not in the span of {a, b}, which means that it cannot be written as a linear combination of a and b. You can prove that {a, b, c} is linearly independent by showing that the only way you can make the linear combination 0 is if all the coefficients are 0. That is, if c1a + c2b + c3c = 0, then c1 = c2 = c3 = 0. For question 2, the statement implies that for each vector vi in T, there exists some scalars a1, a2, ..., ak such that vi = a1u1 + a2u2 + ... + akuk. To show that w is a linear combination of the vectors in S, you need to show that there exist scalars b1, b2, ..., bm such that w = b1u1 + b2u2 + ... + bkuk. This can be shown by substituting the expression for vi into the expression for w, giving you an equation in terms of the scalars a1, a2, ..., ak. Solving this equation will give you the scalars b1, b2, ..., bm that you need.
 


Hello there! It's great that you're looking to understand the proof for linear independence. Let me try to guide you through the process:

1. When we say that {a,b} is linearly independent, it means that these two vectors cannot be written as a linear combination of each other. In other words, there is no way to find a non-zero value for c1 and c2 such that c1a + c2b = 0. This is the definition of linear independence.

Now, if c is not in span {a,b}, it means that c cannot be written as a linear combination of a and b. In other words, there is no way to find non-zero values for c1 and c2 such that c1a + c2b = c. This is because c does not belong to the span of {a,b}.

To show that {a,b,c} is linearly independent, you need to show that there is no way to find non-zero values for c1, c2, and c3 such that c1a + c2b + c3c = 0. This can be done by contradiction - assume that there exist non-zero values for c1, c2, and c3 that satisfy the equation, and then show that this leads to a contradiction (i.e. c belongs to the span of {a,b}).

2. The first sentence means that the set T is made up of vectors that are linear combinations of the vectors in S. In other words, each vector in T can be written as a linear combination of the vectors in S.

To show that w is a linear combination of the vectors in S, you need to find values for b1, b2, ..., bm such that w = b1u1 + b2u2 + ... + bmum. This can be done by using the fact that each vi is a linear combination of the vectors in S. For example, if v1 = a1u1 + a2u2 + ... + akuk, then we can choose b1 = a1 and b2 = a2, and so on. This will give us the desired linear combination of the vectors in S to represent w.

I hope this helps to clarify the essence of the proof for you. Remember, the key is to understand the definitions and use them to guide your reasoning. Best of luck with your studies!
 

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