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If an orthogonal set is linearly independent how can we get the linear combination of these vectors to form another vector that is in the orthogonal set? I thought linear independence meant you cant form another vector that is in the set based off of the vectors in a set.

The only thing that can equal $c_1v_1+c_2v_2+...+c_nv_n=0$ is where scalars $c_1,c_2,...,c_n$ are $0$, meaning no multiples of those vectors can form another vector.

Edit:

For example S=(v1,v2,v3) where v1=[1,2,3] v2=[1,1,-1] v3=[5,-4,1] is an orthogonal subset of R^3.

Vectors in S are non zero so S is linearly indepdent.

Let u=[3,2,1], now we can write u as a linear combination of the vectors in S.

u=c1*v1+c2*v2+c3*v3

Using the equation ci=(dot product of u and vi)/||vi||^2 we can obtain scalars for c where u=c1*v1+c2*v2+c3*v3

Again how can you find this vector u using the linear combination of the vectors in S if S is linearly independent? There should not be a vector u that is a linear combination of the vectors in S if S is linearly independent.

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The definition of linear independence says you can't make 0 out of a linear combination. It says nothing about not being able to make any other vector out of linear combinations.

(1,0) and (0,1) are independent since you cannot write (0,0) = c(1,0) + d(0,1) without c=d=0. But you can write every other vector as a nontrivial linear combination of these. (2,3) = 2(1,0)+3(0,1) for example. Spend some time making sense of the definitions with some concrete examples like this one and it will make sense eventually.

If you call your orthogonal set $\{v_1, v_2, \dots, v_n\}$, you can trivially write any vector in your set as a linear combination (take all coefficients $0$ except the coefficient of $v_k$ which is $1$).

$v_k = 0\cdot v_1+0\cdot v_2+\dots+0\cdot v_{k-1}+1\cdot v_k+0\cdot v_{k+1}+\dots+0\cdot v_n$

This is true of any set, whether it is orthogonal or not.

Moreover, any vector in the span of $\{v_1, v_2, \dots, v_n\}$ can be written as a linear combination of these vectors. This is again true of any set, whether orthogonal or not.

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  • $\begingroup$ but if you make all coefficients except one 0 and we know that all vectors in the orthogonal set are non zero vectors, how does 0*v1+1*v2+...+0*vn=0, which proves that vectors in the set are linearly independent of one another? The concept of linear combination and linear independence being opposites as well as a orthogonal set being linearly independent yet getting the linear combination of the vectors in the set bringing about a new vector in the orthogonal set confuses me. $\endgroup$ – Bob m Nov 21 '14 at 6:03
  • $\begingroup$ It is impossible to have independent vectors sum to $0$, but they can easily sum to one of the vectors in the set (in the way I wrote). $\endgroup$ – RHP Nov 21 '14 at 6:04
  • $\begingroup$ But if they cant sum to zero without the trivial coefficients being 0, doesnt that mean no linear combination of these vectors can form a new vector in the set? $\endgroup$ – Bob m Nov 21 '14 at 6:07
  • $\begingroup$ What do you mean by "a new vector in the set"? You can't write $v_n$ as a combination of $v_1, v_2, \dots, v_{n-1}$ if they are independent. But you can write $v_n$ as a combination of the whole set (in the way I wrote). Also, don't think of "linear combination" as opposite to "linear independence". "Linear dependence" is the opposite. A "linear combination" is a sum of scaled vectors which is used in the definitions of independence and dependence. $\endgroup$ – RHP Nov 21 '14 at 6:09
  • $\begingroup$ Im not sure i understand the difference between the two(v1,v2,vn-1 vs the whole set) $\endgroup$ – Bob m Nov 21 '14 at 6:11

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