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 December 22nd, 2018, 02:19 AM #1 Newbie   Joined: Dec 2018 From: Tel Aviv Posts: 4 Thanks: 0 Feedback for an answer involving Span, Dimension and Linear Independence Hi guys, I have worked out an answer to the following question, but it's stretching my understanding of Linear Algebra to the limit, so… if anybody could have a look at it, and just tell me whether my reasoning is sound, it would be a great help... Let v_1,…,v_k,u,w be vectors in the linear space V. It is given that the equation x_1·v_1+…x_k·v_k=u has a single solution and that the equation x_1·v_1+…+x_k·v_k=w has no solution. a. Prove that the group {v_1,…,v_k} is linearly independent. b. Find the dimension of Sp{v_1,…,v_k,w}. My answer: a. According to the information provided, the equation x_1·v_1+…x_k·v_k=u has a single solution. That means that for the solution to hold, all variables in this equation must have a fixed value, with no free variables. If u is the zero vector, each one of the variables must take on the value 0 for the single solution to hold: it is a given that there is a single solution only, so in this case it must be the trivial solution, as the trivial solution always exists in a homogeneous system. If the trivial solution is the only solution, then v_1 to v_k is linearly independent. b. per definition {v_1,…,v_k} spans Sp{v_1,…,v_k}, and it was shown that {v_1,…,v_k} is linearly independent, so {v_1,…,v_k} is a basis with dimension k. Sp{v_1,…,v_k,w} has an additional member, w, which according to the information given is not a linear combination of {v_1,…,v_k} (given that the equation x_1·v_1+…+x_k·v_k=w has no solution). Therefore {v_1,…,v_k,w} is a basis of Sp{v_1,…,v_k,w}, which dimension k+1. Last edited by skipjack; December 22nd, 2018 at 05:12 PM.
 January 27th, 2019, 06:52 PM #2 Newbie   Joined: Jan 2019 From: Adelaide Australia Posts: 3 Thanks: 0 Hi pregunto, Here's how I would approach the first problem: We are given that x_1 * v_1 + ... + x_n * v_n = u has a unique solution. Let's call this unique solution b = (b_1, ..., b_n). Therefore, b_1 * v_1 + ... + b_n * v_n = u. .... (1) Now we'll prove that {v_1, ..., v_n} is linearly independent using proof by contradiction. That is, we'll assume that it is NOT linearly independent and from that derive something that's impossible. So let's assume that it is NOT linearly independent. In other words, it is a linearly dependent set of vectors. By definition of linear dependence, this means that there exists a vector a = (a_1, ..., a_n) such that a_1 * v_1 + ... + a_n * v_n = 0 (the zero vector) .... (2) Adding equations (1) and (2), we get: (a_1 + b_1) * v_1 + ... + (a_n + b_n) * v_n = u + 0 = u. Therefore, the vector a+b = (a_1 + b_1, ..., a_n + b_n) is another vector, which is distinct from the vector b, which is a solution to x_1 * v_1 + ... + x_n * v_n = u. But this is not possible since we know that that equation has a UNIQUE solution. Therefore, our assumption has lead to an impossibility and hence we conclude that it is an incorrect assumption. Please let me know if this was helpful. Cheers,

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