hello everyone!i'm breaking my head and desperate how to do it...
If i have a vector in linear space V:r v1,v2.....vk,a,b
having A={v1,v2.....vk,a} B={v1,v2....vk,b} C={v1,v2...vk}
and its known that V=sp(A) and b∉sp(C)........
needed to prove
1) a ∈ sp(B)
2) V=span(B)
3) if C not linear depended-->A is basis for V??
thank you very much..(generaly i has a clue but has no idea how its correctly writing the mathematical proof..)
To prove the given statements, we will use the definitions and properties of vector spaces and subspaces. Here's how you can approach each of the proofs:
1) To prove that a ∈ sp(B), we need to show that a can be expressed as a linear combination of vectors in B. Since V = sp(A), we know that every vector in V can be written as a linear combination of vectors in A. So, a can be expressed as a linear combination of vectors in A: a = c1v1 + c2v2 + ... + ckvk + ca.
Now, consider the vector b. We are given that b ∉ sp(C), which means that b cannot be expressed as a linear combination of vectors in C. However, since B contains v1, v2, ..., vk, and b, we can write b as a linear combination of vectors in B: b = c1v1 + c2v2 + ... + ckvk + cb.
Combining both equations, we have a = c1v1 + c2v2 + ... + ckvk + ca = c1v1 + c2v2 + ... + ckb + ca. This shows that a can be expressed as a linear combination of vectors in B, proving a ∈ sp(B).
2) To prove that V = span(B), we need to show that every vector in V can be written as a linear combination of vectors in B, and vice versa.
Since V = sp(A), every vector in V can be expressed as a linear combination of vectors in A. But we have already shown in the previous proof (proof 1) that every vector in A can be expressed as a linear combination of vectors in B. Therefore, every vector in V can be expressed as a linear combination of vectors in B.
Next, we need to show that every vector in B can be written as a linear combination of vectors in V. Since V = sp(A), every vector in V can be expressed as a linear combination of vectors in A. Since every vector in A can be expressed as a linear combination of vectors in B, it follows that every vector in B can be expressed as a linear combination of vectors in V.
By proving both directions, we conclude that V = span(B).
3) To prove that if C is linearly independent, then A is a basis for V, we need to show that A spans V and A is linearly independent.
First, let's show that A spans V. Since V = sp(A), every vector in V can be written as a linear combination of vectors in A. Thus, A spans V.
Next, we need to show that A is linearly independent. Suppose there exists a non-trivial linear combination of vectors in A that equals the zero vector: c1v1 + c2v2 + ... + ckvk + ca = 0, where not all the coefficients c1, c2, ..., ck, ca are zero. Since C is linearly independent, the only way this equation can hold is if ca = 0 and c1 = c2 = ... = ck = 0. But if ca = 0, it means that a = 0, and if c1 = c2 = ... = ck = 0, it means that the linear combination is trivial. Therefore, A is linearly independent.
Since A spans V and A is linearly independent, we conclude that A is a basis for V.
Remember to provide clear reasoning and use mathematical notation to make your proofs more precise and rigorous.