How do you find the basis for the spanning set which contains matrices?

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    Basis Matrices Set
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Discussion Overview

The discussion revolves around finding a basis for a spanning set of matrices. Participants explore the concepts of linear independence and dependence in the context of matrix sets, with a focus on identifying the largest independent subset that can serve as a basis.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • Some participants suggest that not all matrices in the provided link are independent and propose finding the largest independent set as the basis.
  • One participant explains that a basis can be determined by identifying a subset of vectors that spans a vector space and contains only independent vectors, referencing the condition for linear dependence.
  • Another participant provides a specific equation derived from the matrices to illustrate the process of checking for independence and suggests that if the only solution is the trivial one, the matrices are independent.
  • There is a concern expressed regarding the tendency of some users to seek quick answers rather than engaging deeply with the material, which may affect the quality of discussion.

Areas of Agreement / Disagreement

Participants express differing views on the engagement level of users in the forum, with some suggesting that many seek only answers without understanding, while others focus on the technical aspects of finding a basis for the matrices. No consensus is reached on the best approach to the problem or the nature of user engagement.

Contextual Notes

Participants reference specific equations and conditions for linear independence but do not resolve the mathematical steps or assumptions involved in determining the basis for the matrices.

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Given a set of vectors that spans a vector space, the largest subset that contains only independent vectors is a basis. The condition for dependence is that
[tex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0[/tex]
with at least on of the scalars non-zero. If it is, say [\itex]a_i\ne 0[/itex], we can solve for the vector [itex]v_i[/itex] as a function of the others and so can drop it from the set.

Here, that equation is
[tex]a_1\begin{bmatrix}0& 3\\ 1 & 1\end{bmatrix}+ a_2\begin{bmatrix}4 & 5 \\ 3 & 1 \end{bmatrix}+ a_3\begin{bmatrix}-4 & 1 \\ -1 & 1\end{bmatrix}+ a_4\begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}[/tex]
[tex]= \begin{bmatrix}4a_2- 4a_3+ 2a_4 & 3a_1+ 5a_2+ a_3+ a_4 \\ a_1+ 3a_2- a_3+ a_4 & a_1+ a_2+ a_3+ 2a_4\end{bmatrix}[/tex]
which is the same as the four equations
[tex]4a_2- 4a_3+ 2a_4= 0[/itex]<br /> [tex]3a_1+ 5a_2+ a_3+ a_4= 0[/tex]<br /> [tex]a_1+ 3a_2- a_3+ a_4= 0[/tex]<br /> [tex]a_1+ a_2+ a_3+ 2a_4= 0[/tex]<br /> <br /> An obvious solution is [itex]a_1= a_2= a_3= a_4= 0[/itex]. If that is the <b>only</b> solution, the four matrices are independent and so are a basis for there span. If there is a solution in which one of the coefficients is non-zero, we can solve for that matrix in terms of the other three and so drop it from the set.[/tex]
 
@HallsofIvy: I suspect that lawnmowjob asked this as a homework problem, then either figured it out or got help from somewhere else. Given the shortness of his original post, I doubt that he'll be back. It happens too often in these forums, and makes me reluctant to write out more explicit and helpful discussions like yours.

Anyway, I hid the answer in the middle full stop of my answer.
 
Cute! I'll have to remember that!
 
sad, but true. many people just want "the answer", and have little interest in gaining the knowledge.
 

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