Question

Let S=span {v1,v2,v3}, where v1=(1,0,-1,0), v2=(0,1,1,1), v3=(5,4,-1,4). Find the dimension of S and find a basis for S.

Answer 1

If i put those vectors in a matrix and reduce I get
\[\left(\begin{matrix}x _{1} \\ x _{2}\\x _{3} \end{matrix}\right)=x _{3}\left(\begin{matrix}-5 \\ -4\\ 1\end{matrix}\right)\]

Answer 2

what does this tell me???

Answer 3

I'm assuming you were trying to find if they were linearly independent by finding the linear combination that gives the zero vector? Then this means that your 2nd vector can be written as a linear combination of the other 2 vectors. This means that you can remove the 2nd vector from your set and it may still be linearly dependent. To verify this, you repeat the same step as you did above and you should find a unique solution. This will imply that your vectors are linearly independent, so thus they can from a basis (because clearly they are spanning S.

The dimension of S is just the number of vectors in your basis. In this case, there are 2 vectors that form a basis for S, so the dimension = 2

Answer 4

Oops, I should be referring to the 3rd vector, not the 2nd... your equation says x3. Sorry my screen is very small.

Answer 5

so is the basis in R4 or in R3

Answer 6

and what you're saying is to eliminate v3, and just set up the first two vectors as a 4x2 matrix, reduce, and then the resulting vectors will be the basis for S?

Answer 7

Well your basis is a subspace S of R4. R4 refers to the number of components of the vectors in your set, not the number of vectors in the set. So, you have 2 basis vectors. These vectors span the susbspace S. But because there are only 2 vectors, they cannot span all of R4.

Answer 8

And yes to your question. There's a theorem that says smtg like "Let V be a vector space with v1, ..., vn are in V. Then if vn is in Span(v1, ..., v_n-1}, then Span(v1, ..., vn} = Span{v1, ..., v_n-1}

Basically this says that if your vector space has K vectors and you have an extra vector (K+1) that is in the span, then you can simply remove the K+1 vector and you still have the same spanning set.

Answer 9

well if i reduce just the first two vectors i get
(1,0,0,0) and (0,1,0,0)
and you're saying that is the basis for S? and since it has two columns, or bc it has to pivot columns that the dim is 2

Answer 10

if you get(1,0,0,0) and (0,1,0,0) this implies x1 = 0 and x2 = 0 is the ONLY solution to the system of equations. Thus, we cannot write the vectors v1 and v2 as linear combinations of each other so they are linearly independent. You have to write the original vectors v1 and v2 [1,0,-1,0] and [0,1,1,1] as your basis. The (1,0,0,0) and (0,1,0,0) are just the row-reduced versions that helped you find the solution to your system of equations.

We know this is a basis because... a basis means:
1) the vectors are linearly independent (we just verified this)
2) the vectors span the set (here, we are verifying they span the subspace S). We know this is true because S = Span {v1, v2, v3}. We found out we could eliminator v3, so by the Theorem I showed you, S = Span(v1, v2}. So v1, v2 span S

Answer 11

so would i have to show that x3 can be written as a linear combination of the others, before i am allowed to eliminate it?
then after eliminating it i have to show that v1 and v2 are independent without v3, and since they are keep them as the basis for S, and since there are two vectors in the basis the dim is 2.

Answer 12

Yes, theoretically you should show x3 can be written as a linear combination of x1 and x2. But it turns out that every time you get a free variable when row-reducing, the free variables can always be written as linear combinations of the the other non-free variables, and thus you can eliminate them from your span.

And yes after you have v1, v2 you show their linear independence and thus you will keep them as your basis. And so dim = 2

Answer 13

But it's very easy to show that x3 is a linear combination of x1 and x2.
x3 = 5x1 + 4x2

Answer 14

thank you for your help... appreciate it.

Answer 15

no problem :)