Linear algebra: basis

[_Bd_]

Homework Statement

find the basis of a subspace of R^3 spanned by S:

1. S = { (4,4,8) (1,1,2) (1,1,1)}
2. S = { (1,2,2) (-1,0,0) (1,1,1)

Homework Equations

Im allowed to use calculator.

The Attempt at a Solution

Im not really sure what this is about. . .I tried the following and got the correct answer on a previous problem. . .but this time it didnt work:

first reduce the matrix to row echelon form which on problem 1 is:
[1 1 2]
[0 0 1]
[0 0 0]
therefore its rank 2 and the basis should be (1, 1, 2) (0, 0, 1) ?
the answer in the back is (1, 1, 0) (0, 0, 1)
I have like no clue as to what to do to get there. . .
so for the second one following the same steps I get
[1 2 2]
[0 1 1]
[0 0 0]
therefore again its rank 2 and . . .well I don't know how to get the basis? trying a guess Id say its (0, 0, 1) and (1, 1, 0)
?? I mean what is that based on? could it maybe be (1, 0, 1) (0, 1, 0) also?



I already re-read the book and just don't understand it. . .
tho people recommended me to not understand linear algebra (yet) and just follow steps. . .but still how do I get there?

-thanks

 

[radou]

What the people recommended is complete nonsense. This is always much easier if you think about it.

For example, in your first set of vectors, is there one which is a linear combination of the others (or another one)?

 

[benorin]

On the first one: if $$R_{i}$$ is the i^th row, then puit $$R_{1}-2R_{2}\rightarrow R_{1}$$

 

[HallsofIvy]

A basis for a vector space has two properties- it spans the space and its vectors are linearly independent. You are given that these vectors span the space. Are they linearly independent? If so the entire set is already a basis. If not, then one of them can be written as a linear combination of the others and can be dropped from the set. Now check to see if this new set is linearly independent. If so, it is a basis. If not, you can drop another vector.

For (1) you might note that (4, 4, 8)= 4(1, 1, 2).

For (2), (1,2,2)= 2(1,1,1)+ (-1,0,0).

 

[_Bd_]

A basis for a vector space has two properties- it spans the space and its vectors are linearly independent. You are given that these vectors span the space. Are they linearly independent? If so the entire set is already a basis. If not, then one of them can be written as a linear combination of the others and can be dropped from the set. Now check to see if this new set is linearly independent. If so, it is a basis. If not, you can drop another vector.

For (1) you might note that (4, 4, 8)= 4(1, 1, 2).

For (2), (1,2,2)= 2(1,1,1)+ (-1,0,0).

so 1 of each is dropped. . .therefore they are both rank 2.
the thing is, once I determine that they are rank 2 how do I write the basis, I thought it would just be the remaining two vectors (as you said the set would be a basis)
so (according to the book) the first thing is to reduce the matrix to echelon form afterwards you can see if any row is made of zeros (then you can eliminate it as you mentioned since its dependent) the other 2 should form the basis (since they are already reduced) but the answer I got was wrong and that's what I was wondering.

@radou
I think the same way (that its better to understand it) but seriously throughout this semester I've been making it by just following instructions and so far tho I know processes I have no clue as to what they mean or how to apply them. . . I mean I get like the gist of it (most mathermatical terms like linear independence and other stuff) but as to its appliications and stuff most of the tihngs I don't get.

anyways. . .so I already got that both are rank 2 because 1 of the vectors is a linear combination of the other 2 (therefore the ref(A) has a row of zeros). . .but what about the basis? I still don't get it (I mean how to get to the answer in the back of the book)