Definition talk:Nondegenerate Tuple of Elements of Scalar Product Space

Comment to User:Usagiop
The formulation could be clearer. I would write the requirement as:


 * Suppose for each $j \in \set {1, \ldots, k}$ and for all $i_1, \ldots, i_j \in \set {1, \ldots, k}$ with $i_1 < i_2 < \ldots < i_j$ that:
 * the vectors $\tuple {v_{i_1}, \ldots, v_{i_j} }$ span a nondegenerate $j$-dimensional subspace of $V$.

However, I do not have access to the cited source, so I cannot be certain what the actual definition is. --Anghel (talk) 14:44, 27 January 2023 (UTC)


 * Thank you for pointing out my mistake. You are right. I improved my comment, since this definition seems still wrong. I guess the original definitions should require any $j$-subtuple of the given $k$-tuple to span a $j$-dimensional degenerate space. Someone should check it. --Usagiop (talk) 15:11, 27 January 2023 (UTC)


 * Sorry you have already written the same. --Usagiop (talk) 15:18, 27 January 2023 (UTC)


 * Now I think again it should be that $\tuple {v_1,\ldots,v_j}$ for all $1\le j \le n$ spans a $j$-dim degenerate space. I am waiting for more input here. --Usagiop (talk) 15:30, 27 January 2023 (UTC)


 * We have that the $j$-dimensional space should be nondegenerate with respect to the scalar product $q$. Unfortunately, the original author does not define scalar product, but from their link on the Definition:Scalar Product Space page, $q$ should be defined as a: Definition:Nondegenerate Symmetric Bilinear Form. That is, the difference from a real inner product is that $q$ is not positive-definite, but nondegenerate.


 * As an example, let $V := \span \set{ e_1,e_2}$ be a $2$-dimensional vector space over $\R$. Define a scalar product $q:V \times V \to \R$ by:
 * for $x_1, x_2,y_1,y_2 \in \R$ : $\map q {x_1 e_1 + x_2 e_2, y_1 e_1 + y_2 e_2} = y_1 x_1 + y_1 x_2 + y_2 x_1$


 * Then $q$ is symmetric and bilinear (but of course, not positive-definite). Also, $q$ is nondegenerate: There is no $u \in V \setminus \set{\bszero}$ such that for all $v \in V$ : $\map q {u,v} = \bszero$.


 * Then, $\span \set {e_1}$ is a nondegenerate subspace. However, $\span \set {e_2}$ is degenerate, as:
 * for all $x_2, y_2 \in \R$ : $\map q { x_2 e_2, y_2 e_2 } = 0$


 * $V = \span \set{e_1,e_2}$ is nondegenerate, but should $\set {e_1,e_2}$ be defined as a Definition:Nondegenerate Tuple of Elements of Scalar Product Space? I believe the intention is that they should not. So, you should require more that $\tuple {e_1,\ldots,e_j}$ for all $1\le j \le n$ spans a $j$-dim nondegenerate space. --Anghel (talk) 15:57, 28 January 2023 (UTC)


 * Sorry, why is your $q$ symmetric? $\map q {e_1, e_2} = 2 \ne 3 = \map q {e_2, e_1}$? --Usagiop (talk) 16:27, 28 January 2023 (UTC)


 * My fault, there was a typo in the definition of $q$. Has been corrected now to:
 * $\map q {x_1 e_1 + x_2 e_2, y_1 e_1 + y_2 e_2} = y_1 x_1 + y_1 x_2 + y_2 x_1$
 * --Anghel (talk) 20:17, 28 January 2023 (UTC)


 * OK, I exactly meant this should be the case. I believe the definition just requires that there exists a nondegenerate sequence $V_1 \subset V_2 \subset \cdots \subset V_n = V$ such that $\dim V_j = j$. But, of course, I do not know the author's intention. --Usagiop (talk) 21:49, 28 January 2023 (UTC)


 * Your definition does agree with the source. I found the source book on Google books. There are a few pages visible, and luckily, the nondegenerate definition can be seen. So you are correct.--Anghel (talk) 22:26, 28 January 2023 (UTC)
 * Should be done. --Usagiop (talk) 23:20, 28 January 2023 (UTC)