Talk:Linear independence

message: It should at least be mentioned that in a module over a ring, linear independence (in the sense that zero cannot be represented as a non-trivial linear combination) is not equivalent to one vector being contained in the span of the remaining elements.--80.136.131.201 18:37, 20 January 2006 (UTC)
replies: text: This article is only about linear independence over the real numbers. For general rings, things are messy indeed, and I don't even think that the concept if linear independence is at all useful. Oleg Alexandrov (talk) 23:45, 20 January 2006 (UTC)
replies: text:  Some parts of the usual theory work pretty well: Linear independence of the columns of a matrix corresponds to uniqueness of solutions (if any exist), a subset of a module is a basis iff it is linearly independent and generating. Cf. the characterization based on properties of the induced map K^n\to V mentioned by KSmrq in the preceding section.--80.136.131.201 00:14, 21 January 2006 (UTC)
thread_title:  Linear independence over rings 

message: The article states that a check of independence can use the determinant when the number of vectors equals the dimension. In effect, this is a check to see if the vectors form a basis. But even when the number of vectors, m, is less than the dimension, n, we can check for independence by requiring all m×m subdeterminants to be nonzero. For example, the following three vectors of dimension four are independent.
:v_1 = \begin{bmatrix}4\\5\\6\\7\end{bmatrix},\quad v_2 = \begin{bmatrix}2\\3\\5\\8\end{bmatrix},\quad v_3 = \begin{bmatrix}1\\1\\2\\3\end{bmatrix}
:\begin{vmatrix}4&2&1\\5&3&1\\6&5&2\end{vmatrix} \ne 0,\quad \begin{vmatrix}4&2&1\\5&3&1\\7&8&3\end{vmatrix} \ne 0,\quad \begin{vmatrix}4&2&1\\6&5&2\\7&8&3\end{vmatrix} \ne 0,\quad \begin{vmatrix}5&3&1\\6&5&2\\7&8&3\end{vmatrix} \ne 0
I'd suggest this is worth mentioning in the article. --KSmrqT 01:15, 31 January 2006 (UTC)
replies: text: Sure, put it in.  Personally, I always like to err on the side of exhaustiveness, this being an encyclopedia, not a textbook.  By the way, both statements are implied by the more high-brow statement that a set of vectors is dependent iff their wedge product is 0.  -lethe talk + 01:54, 31 January 2006 (UTC)

text: Shouldn't that be 'by requiring at least one of the sub-determinants to be non-zero'? 137.205.139.149 20:58, 19 January 2007 (UTC)
replies: text:  You're a year late to the party. The article has already been augmented, and you can check that the language used there is correct. It says we have dependence (not independence) if all subdeterminants are zero, which (as you correctly observe) means that at least one nonzero subdeterminant indicates independence. But as a practical matter, for numerical computation we would use a rank-revealing decomposition. --KSmrqT 17:12, 20 January 2007 (UTC)
replies: text:  I disagree. I was reading the article and it wasn't clear to me whether it was required that all determinants be zero or one of them being zero was enough to have dependence. Maybe the language is correct, but it is not clear enough for those of us trying to learn.  — Preceding unsigned comment added by Viviannevilar (talk • contribs) 02:14, 22 June 2011 (UTC)
thread_title:  determinant check 

message: Did anyone actually solve the first example? I think it should be (9,2,3) instead of (4,2,3).  —Preceding unsigned comment added by Caelumluna (talk • contribs) 02:38, 20 September 2007 (UTC)
replies: text: It looks fine to me. -- Jitse Niesen (talk) 04:30, 20 September 2007 (UTC)
thread_title:  Example in lead 

message: Hi, I could be wrong here but I think the first line in the article needs to be rewritten. It reads:
In linear algebra, a family of vectors is linearly independent if none of them can be written as a linear combination of finitely many other vectors in the family
Strictly speaking, shouldn't it instead read something like ...if no linear combination of finitely many vectors in the family can equal zero.
In a vector space over Z2, for example, (2,0) and (3,0) would incorrectly be classed as linearly independent by the first definition, since neither is an integer multiple of the other.  — Preceding unsigned comment added by Insperatum (talk • contribs) 00:41, 17 October 2012 (UTC)
replies: text: There is no vector space over Z, the integers, because Z is not a field. Your suggested wording introduces complications (and the existing wording does not have the problem you suggest for the reason I've given): the sum of a no vectors (zero is a finite number) equals zero, making every set "linearly dependent" by your definition. — Quondum 04:18, 17 October 2012 (UTC)

text: Hi. the first line at the moment "In the theory of vector spaces, a set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the others; if no vector in the set can be written in this way, then the vectors are said to be linearly independent. " doesn't make it apparent that {0} is a linear dependent set as there are no other vectors in the set.  — Preceding unsigned comment added by 146.179.206.233 (talk) 12:42, 25 November 2016 (UTC)
thread_title:  Inaccuracy in first line? 

message: It seems to me that the algebra is incorrect in the $R^{2}$ section (the 2-dimensional section).  The row reduction should not create a unit vector... it looks like a math error.  But I didn't want to change it, in case I'm wrong.  Can someone verify that it's incorrect?  (It should give a matrix with the first column as [1, 0], which is correct, but the second column should be [0, 5], shouldn't it?
This.is.mvw (talk) 22:42, 1 August 2017 (UTC)
replies: text: I see nothing wrong with the computation as it is given. The algebraic steps are exactly those given above in the three vector case. The second columns are respectively, [-3, 2], [-3, 5], [-3, 1], and [0, 1]. --Bill Cherowitzo  (talk) 02:44, 2 August 2017 (UTC)

text: I find it interesting that the latest revision is designed to include a definition for the linear dependence of sets consisting of zero and one vectors.  Can the empty set of vectors really be contemplated as being linearly independent or linearly dependent?  I think it is even difficult to view a single vector as linearly dependent or independent on its own.Prof McCarthy (talk) 22:41, 4 September 2018 (UTC)
thread_title:  Evaluating Linear Independence