Basis for Subspace Spanning Three Vectors

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    Basis Subspace
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Discussion Overview

The discussion revolves around determining the basis of the subspace spanned by three vectors, specifically whether to consider the basis of the row space or the column space of a matrix formed by these vectors. Participants explore the implications of row reducing the matrix and the correctness of their approaches in a testing context.

Discussion Character

  • Debate/contested
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant asks whether the basis of the subspace spanned by three vectors is found from the row space or the column space after forming a matrix.
  • Another participant suggests that if one starts with rows and row reduces, the basis corresponds to the row space.
  • A participant recalls a test experience where they wrote the vectors as columns and found the basis for the column space, which was marked wrong, leading to confusion about the correct approach.
  • Some participants argue that if column operations are performed correctly, the answers should be equivalent, but the grading may reflect a misunderstanding of the initial setup.
  • There is a discussion about the process of finding the basis for the column space, emphasizing the importance of identifying leading 1's after row operations.
  • A participant describes their method of arranging vectors as columns, performing row operations, and returning to the original columns, questioning if this approach was valid.
  • Another participant provides a simple example to illustrate the concept of spanning and suggests that the operations should not lead to confusion if done correctly.
  • A participant shares a specific test question and their solution, asserting confidence in their algebra and the correctness of their answer.
  • One participant expresses skepticism about the grading and suggests consulting the TA or professor for clarification.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of their methods for finding the basis of the subspace, with some uncertainty about grading practices and the interpretation of operations performed on the matrix. No consensus is reached regarding the best approach or the implications of the grading outcomes.

Contextual Notes

Participants highlight potential confusion arising from the distinction between row and column operations, as well as the importance of clearly stating the method used in tests. There are references to specific grading experiences that may not reflect a universal standard.

EvLer
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Hi everone,
quick question: if i have 3 vectors and I need to know basis of the subspace they span, so I write each vector as rows of a matrix and when I reduce it, is it the basis of rowspace or column space of matrix that is the basis of the subspace that vectors span?

Thanks in advance.
 
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As you start by writing them as rows and then row reduce (hopefully), want to havae a wild guess?
 
basis for the rowspace?

edit:
on a test one time I wrote them as colums and found the basis for columnspace -- it was marked wrong, so now I'd want to be sure to know this.
 
as long as you declared you were writing them as columns and did colmun operations, csurely you can see that the answers would be the same? if it wasn't clear that was what you were doing then i can see why they might mark it wrongly. more likely is that you got a different (but equivalent) answer from the mark scheme, the marker looked at the working to see if it was valid and noticed your initial matrix was wrong and gave you no more marks. this is common as unfortunate as it may be since no one actually has patience to accurately mark work as tedious as this in great detail.
 
EvLer said:
on a test one time I wrote them as colums and found the basis for columnspace -- it was marked wrong, so now I'd want to be sure to know this.

How did you find the basis for the columnspace? Maybe you did this incorrectly? If you used row operations to reduce it, you find the columns with leading 1's. The corresponding columns in the original matrix make the basis for the comlumn space. The bold part is a common mistake here.
 
thanks for the replies :smile:

shmoe said:
How did you find the basis for the columnspace? Maybe you did this incorrectly? If you used row operations to reduce it, you find the columns with leading 1's. The corresponding columns in the original matrix make the basis for the comlumn space. The bold part is a common mistake here.
What i did was arrange all vectors as colums of a matrix. I did use row operations to reduce it and then went back to the columns of the original matrix, THAT part I did remember. So, was it wrong to do row operations in this case to find a basis of the column space? (edit: that is what I am confused about)
 
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yes. think of a simple example. (1,1,1) and (1,0,0) clearly the span is the 2 d plane spanned by (1,0,0,) and ((0,1,1) try putting them as columns then doing row ops.
 
EvLer said:
What i did was arrange all vectors as colums of a matrix. I did use row operations to reduce it and then went back to the columns of the original matrix, THAT part I did remember. So, was it wrong to do row operations in this case to find a basis of the column space? (edit: that is what I am confused about)

As long as you go back to the original columns before doing the operations (and didn't make any algebra mistakes), it's good.

There's the danger something here is getting lost in translation though, if you have your test handy would you like to post the question and your answer? If it's wrong, we can work through what you did in more detail. (Even if it's right we can work through it in more detail if you like too)
 
OK, here it is:

find a basis for the subspace spanned by the vectors
{(1, -1, 2), (5, -4, 1), (7, -5, -4)}

so I reduced the matrix which consisted of these vectors arranged as columns and after reduction, I had 2 pivots in first and second column, so my answer was {(1, -1, 2), (5, -4, 1)} and I'm pretty sure I did not make algebra mistakes.
Oh and the note from grader said: this is the column space.
 
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  • #10
Assuming no algebra mistakes (ruling out a 'fluke'), what you've done is good.

No matter how you found your answer, it's correct. The third vector is a linear combination of the first two, and the first two vectors are linearly independent.

I hate to second guess someone, but TA's sometimes make mistakes. I'd go ask him or her about it, or your professor.
 
  • #11
Thank you, I will take your advice :)
 

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