Question

Is there a way to check this using the c1,c2,c3 method to see if it spans in R2 {(1,3),(-2,-6),(4,12)}

Answer 1

I can obviously see that vectors 2 and 3 are linear combinations of vector (1,3) but is there a way to show work in this?

Answer 2

you cant just state it in words?

just say:

\[\left(\begin{matrix}-2 \\ -6\end{matrix}\right)=-2\left(\begin{matrix}1 \\ 3\end{matrix}\right)\]

and

\[\left(\begin{matrix}4 \\ 12\end{matrix}\right)=4\left(\begin{matrix}1 \\ 3\end{matrix}\right)\]

Therefore there is a non trivial linear combination of the vectors that gives 0. Namely:

\[1v_1+\left(\frac{1}{2}\right)v_2+0v_3=0\]

Answer 3

are you saying this is a subspace?

Answer 4

Im saying they are linearly dependent, not independent. So they couldnt span R^2

Answer 5

I was looking for a definition in a sense of dependent and indepedent for spans but i couldn't find it. could you give me that?

Answer 6

They are all just multiples of the same vector like you pointed out, so their span is only one dimensional

Answer 7

would it be that if the vectors are linearly independent, then S spans. If dependent, they do not span S?

Answer 8

Thats true to a certain extent. If we are in n-dimesional space, and we have n linearly independent vectors, then yes you are correct.

Answer 9

However, if you have less than n vectors, even if they are linearly independent they can possibly span the n dimensional space.

Answer 10

so only when working within R^n spaces

Answer 11

er, im sry, i mean cant* they cant span the space.

Answer 12

so for example the set (1,-2,0),(0,0,1),(-1,2,0) the set doesn't span R^3 because 1,1,0 is within the space and cannot be obtained

Answer 13

due to the linear dependence of the two vectors?

Answer 14

exactly. the last vector is just -1 times the first. so its not adding any new info.

Answer 15

my book says otherwise =/

Answer 16

really? =/ theres no way those three vectors can span R^3. They span a 2 dimensional subspace (all the linear combinations of the first and second vector), but thats it. One way to see if you have linearly independent vectors is to make the the columns of a matrix and row reduce. in this case you you would have a 3 by 3 matrix. if you get the identity, you have a linearly independent set of vectors that span all of R^3. If you dont, they are linearly dependent. The number of rows with pivots (called the Rank of a matrix) tells you the dimension of the subspace spanned by the vectors.

Answer 17

wait nvm it does say it doesn't span but it does it some weird way... can you perhaps figure it out since i believe that is what i need to do... .they let r^3 = u1,u2,u3 and then solve the system equation

Answer 18

like it says infinite if -2u1 or something

Answer 19

thats weird =/ i dont understand what they are trying to do.

Answer 20

they are trying to make a system of equations out of it it looks like so lik

c1(1,-2,0)+c2(0,0,1)+c3(-1,2,0)

Answer 21

=(u1,u2,u3) i just dont see what they do after to assume that it is in the space