# Basis for Subspace Spanning Three Vectors

• EvLer
In summary, the basis for the subspace spanned by the vectors is found by arranging them as columns of a matrix and then reducing it with row operations. If the reduction is done correctly, the first and second vectors in the original matrix will be the pivots for the new matrix.
EvLer
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?

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

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)

Last edited:
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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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 independant.

I hate to second guess someone, but TA's sometimes make mistakes. I'd go ask him or her about it, or your professor.

## What is the basis for a subspace spanning three vectors?

The basis for a subspace spanning three vectors is a set of three linearly independent vectors that can be used to represent any vector within the subspace.

## How do you determine if three vectors can be used as a basis for a subspace?

To determine if three vectors can be used as a basis for a subspace, you can check if the vectors are linearly independent. This means that none of the vectors can be written as a linear combination of the others.

## What does it mean for a set of vectors to span a subspace?

If a set of vectors spans a subspace, it means that any vector within that subspace can be written as a linear combination of the vectors in the set. In other words, the vectors in the set "span" the subspace.

## Why is it important for a subspace to have a basis?

The basis for a subspace is important because it provides a way to represent any vector within the subspace using a linear combination of the basis vectors. This makes it easier to work with and understand the properties of the subspace.

## Can a subspace have more than one basis?

Yes, a subspace can have more than one basis. This is because there can be multiple sets of linearly independent vectors that can be used to represent the same subspace. However, all bases for a subspace will have the same number of vectors.

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