# What means to complete a pair of vectors $(w, s)$ to an arbitrary basis of $R^d$?

I found in an article this : Let $B = (b_1, b_2, . . . , b_d)$ be an orthonormal basis of $R^d$ such that $<b1, b2 >=< w,x >$ (where $< ... >$ denotes linear span). In order to construct $B$, first complete $(w, x)$ to an arbitrary basis of $R^d$ and then apply Gram-Schmidt orthonormalization process.

My question is : what means to complete a pair of vectors $(w, s)$ to an arbitrary basis of $R^d$ ?

I read this how to complete arbitrary basis knowing 2 orthonormal vectors of Rd (d > 2)$2$-orthonormal-vectors-of-rd-d-2

but I didn't understand how to choose randomly these vectors : "Practically, one could choose $d−2$ random vectors $x_1,…,x_{d−2}$ (e.g. be choosing each coordinate following a Gaussian distribution)."

Please use as many details as possible in your explanations.

## 1 Answer

To complete a pair of vectors $(w,s)$ to an arbitrary basis of $\mathbb R^d$ means to find an ordered basis of $\mathbb R^d$ such that $w,s$ are the first two vectors of it. Of course there are a lot of differen possible choices. Not at all completely random however.

The third vector $v_3$ should be chosen as not to belong to $\langle w,s\rangle$.

The (eventual) fourth vector $v_4$ should be chosen as not to belong to $\langle w,s, v_3\rangle$ and so on...

In this way each vector is not a linear combination of the previous ones and this is enough to show that they are all linearly independent. When you've got $d$ of them you have a basis.

Nothing you've done so far guarantees you that this basis is orthonormal. At this point you apply Gram-Schmidt.

• Thank you! Now I've figured out what I have to do. May 26, 2015 at 6:45