If a vector space has an inner product defined on it (i.e. it is an
inner product space), then we may also say that a set of orthogonal vectors
(a set where each is orthogonal to all others) will be linearly independent. In
Euclidean space, this is simply saying that a set of vectors that are all
perpendicular to each other are linearly independent, which should be fairly
intuitive if you think about the examples given in Gorgonzola's
write-up. It is also relatively simple to prove.
In the following I will denote vectors in bold. Assume we have a set of vectors v[i], where i goes from 1 to n,
which are orthogonal with respect to the inner product that I will denote
<a,b>. So, this
means that, for any i and j, where i not equal to j, <v[i],v[j]>=0. Let's assume that none of the v[i] are the
0 vector.
In order to be linearly dependent we require that there exist
a set of scalars l[j], where j goes from 1 to n and at least one of the l[j] is
nonzero, such that
l[1]*v[1]+...+l[n]*v[n]=0
Now we make take the inner product of each side with one of the vectors v[j]
<v[j],l[1]*v[1]+...+l[n]*v[n]>=<v[j],0>
We know that for any vector a, <a,0>=0, so together with the linearity of the
inner product, we have
l[1]*<v[j],v[1]>+...l[n]*<v[j],v[n]>=0
Now, all the inner products here vanish except the one that v[j] with itself
l[j]*<v[j],v[j]>=0
Since, by assumption, none of the vectors v[i] are the 0 vector, then
<v[j].v[j]> must be a non-zero scalar, by the definition of the inner product.
Thus, to make the equation true, it must be that l[j]=0. But as you can see, I
never specified what j was, so this is true for all j from 1 to n. We have proven
that l[j]=0 for all j from 1 to n,
meaning only the "trivial solution" exists; therefore,
the vectors are linearly independent. It's important to note that while this
is intuitively true for the normal, Euclidean inner product, the proof above
holds for any inner product space. We we can also see n must be less than or equal to the dimension of the space, otherwise we would have a set of linearly independent vectors that number greater than the dimension, which is a contradiction.
I just read Gorgonzola's really great write-up above (upvote it! :) ) this morning, and
was moved to write this addendum. Since I just made up this
proof off the top of my head, please let me know if there are
any mistakes. Since it's pretty stright forward I think (and hope) there
aren't.