Question

wat do they mean by this:1.A set{alpha(i)} of non-zero vector is linearly dependent if and only if some (alpha)k is a linear combination of the alpha(j) with j

Answer 1

a linear combination of vectors us a set of vectors added together multiplied by some constant

Answer 2

say we have a set of n vectors\[\{\vec v_1,\vec v_2,...,\vec v_n\}\]if any vector \(\vec v_k\) in the set can be written as\[\vec v_k=c_1\vec v_1+c_2\vec v_2+...+c_n\vec v_n\]where \(c_1,c_2,..c_n\) are scalars and at least two are non-zero

Answer 3

for example \[\vec v_1=\langle1,2,3\rangle\]\[\vec v_2=\langle1,0,3\rangle\]\[\vec v_3=\langle3,2,9\rangle\]are not a set of linearly independent vector because one can be written as a linear combination of the other two\[\vec v_3=\vec v_1+2\vec v_2\]

Answer 4

the notation you have presented is strange, I just trying to illustrate the concept

Answer 5

on the first question wat do they mean by saying j<k

Answer 6

that I'm not too sure about... I think I need to be more familiar with exactly what they mean by k, i, and j in your book

Answer 7

k;i;j are the foot notes

Answer 8

yes, but does i pertain to the number of vectors and j to the number of elements?
is k a random vector in the set (I think that's what they mean by that part)

Answer 9

1.A set \(\{\alpha_i\}\) of non-zero vector is linearly dependent if and only if some \(\alpha_k\) is a linear combination of the \(\alpha_j\) with \(j<k\).

2.A set \(\{\alpha_i\}\) of non-zero vector is linearly independent iff \(\alpha_k\) is not an element of \(\langle\alpha_j,......,\alpha_{k-1}\rangle\)

is this exactly how it is stated?

Answer 10

yes

Answer 11

I feel like I understand linear independence, but this phrasing is confusing me
I'm not sure why they require j<k

Answer 12

@phi

Answer 13

but he is not there .Thanks

Answer 14

sorry, he is online though...
hopefully he'll help us out
good luck!

Answer 15

they are indexing (labeling) each vector.
If you start with vector 1, and start adding vectors, if you get to vector k, and it is linearly dependent on the previous vectors 1 to k-1, then the set is linearly dependent.

It is (obviously?) true. If you get to the end of the all the vectors and none are linearly dependent, then the set is independent.

Answer 16

ok, that makes sense
I just couldn't see the point of arranging the vectors in some order like that, such that the vector in question v_k is at the end of the list.

Answer 17

it makes sense

Answer 18

I'm guessing so you can go on to define a basis (minimum number of vectors that are independent)

Answer 19

such that the vector in question v_k is at the end of the list.

that is a bit misleading. say a max of three vectors can be independent and you start with v1,v2,v3,v4

any of the 4 could be in the independent set e.g. (v4,v3,v2) the last one will always be dependent of the first 3.

Answer 20

does the definition of basis reflects only to the finite sets only?

Answer 21

I only learned what Strang taught in his course.

Answer 22

@jacobian check this