# Linear combination of vectors in orthogonal set

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.

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.

• 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. Nov 21 '14 at 6:03
• 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).
– RHP
Nov 21 '14 at 6:04
• 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? Nov 21 '14 at 6:07
• 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.
– RHP
Nov 21 '14 at 6:09
• Im not sure i understand the difference between the two(v1,v2,vn-1 vs the whole set) Nov 21 '14 at 6:11