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[Linear Algebra] Maximal linear independent subset
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[QUOTE="iJake, post: 6016330, member: 645939"] [h2]Homework Statement [/h2] In the follow cases find a maximal linearly independent subset of set ##A##: (a) ##A = \{(1,0,1,0),(1,1,1,1),(0,1,0,1),(2,0,-1,)\} \in \mathbb{R}^4## (b) ##A = \{x^2, x^2-x+1, 2x-2, 3\} \in \mathbb{k}[x]## [h2]The Attempt at a Solution[/h2] The first part of the exercise is trivial, as it is easy to observe that the second vector is a linear combination of the first and third vectors. My question is one of mechanics. Should I be writing the elements of each vector as the rows or columns of a matrix? In a previous exercise, I had to determine whether or not a vector space was spanned by a set of vectors. In that case, it was: Determine if ## V = \mathbb{k}[x]_3## is spanned by ##A = \{1, 1+x^2, 1-x+x^2+x^3, 4-x+2x^2+x^3\}## I reordered the elements and presented them as the columns of a matrix: ## \begin{matrix} 0 & 0 & 1 & 1\\ 0 & 1 & 1 & 2\\ 0 & 0 & -1 & -1\\ 1 & 1 & 1 & 4\\ \end{matrix} ## I did row operations (switch R4 with R1, then with R3, add R3 to R4) and found a row of 0s, thus telling me the rank of the matrix is 3 and that A does not span V. However, for my current problem, I wrote the matrix using the vector elements as rows, not columns. When I write them as columns I do not reach the same conclusion, which confuses me. ## \begin{matrix} 1 & 0 & 1 & 0\\ 1 & 1 & 1 & 1\\ 0 & 1 & 0 & 1\\ 2 & 0 & -1 & 0\\ \end{matrix} ## Here I clearly observe that the second row is the sum of the first and third rows. However, if I write it where the vector elements are columns, I see something else: ## \begin{matrix} 1 & 1 & 0 & 2\\ 0 & 1 & 1 & 0\\ 1 & 1 & 0 & -1\\ 0 & 1 & 1 & 0\\ \end{matrix} ## Can someone explain the difference to me? Thanks. [/QUOTE]
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[Linear Algebra] Maximal linear independent subset
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