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.