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It is not always the case that linear independence mean orthogonality, but orthogonality implies linear independence in real and complex vector space. Is there a vector space over some field where orthogonality does not imply linear independence?

Zee
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    Inner products are conventionally only defined over $\Bbb{R}$ or $\Bbb{C}$. There was a recent question inquiring about how we could extend the notion to other fields, but conventionally, inner products can't exist in spaces over other fields. – user780985 May 01 '20 at 17:40

2 Answers2

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If two non-zero vectors $x_1,x_2$ are orthogonal but linearly dependent, then

$$a_1x_1+a_2x_2=0$$ for non-zero $a_1,\,a_2$. and if we take the inner product of both sides with $x_1$ then we get $$a_1|x_1|^2+a_2\langle x_1,\,x_2\rangle=0$$ which means that $|x_1|=0$, a contradiction.

J.G.
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Paul
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The issue is the original post is attaching the word 'orthogonal' solely to 'inner product' without saying so. Doing so is quite common-- except it contradicts the common usage of 'orthogonal complement' -- see end.

In general we may define orthogonal by $\langle \mathbf w, \mathbf v\rangle = 0$
and we want this to be an equivalence relation so it should imply $\langle \mathbf v, \mathbf w\rangle = 0$

but $\langle , \rangle$ this need not be an inner product -- it could be a symmetric bilinear form that is merely positive semidefinite or even indefinite (e.g. Lorenz Form). In such cases orthogonality doesn't imply linear independence--- this is most easily seen with non-zero null vectors which are orthogonal to everything including themselves.

one may further generalize this to e.g. skew forms, where every vector is self-orthogonal and this holds over arbitrary fields.

re: orthogonal complement
Probably the simplest and most commonly encountered case of 'orthogonal' being decoupled from inner products comes from (implicitly) using the dot product as a bilinear form.

E.g. consider the vector space $\mathbb F_3^n$, where we select $n = k3$ for some natural number $k$
$A:= \mathbf 1\mathbf 1^T$, then we have
$\mathbf A\mathbf 1 = \mathbf 0$

people will commonly say things like $\mathbf 1$ is in the orthogonal complement to the row space of $\mathbf A$, but $\mathbf 1$ is hardly linearly independent from $A$'s rows -- in particular it is exactly equal to any row of $A$.

user8675309
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