# Why is a set containing a single vector orthogonal?

I recently started studying inner products, norms and orthonormal vectors and bases to find the projection of a vector.

I have a problem in this statement: Given, $$V=\mathbb{R}^2$$ and is an inner product space, $$W\subset V$$ and $$W= \text{span}\lbrace (2,1) \rbrace$$, then $$\langle(2,1) \rangle$$ is orthogonal. Here, $$\langle\cdot,\cdot \rangle$$ denote standard inner product which is the dot product.

I understand that for a set $$V=\{v_1,v_2,...,v_k\}$$ to be orthgonal, $$\langle v_i,v_j \rangle=0$$ for $$i\neq j$$. As per me, for $$\langle(2,1) \rangle$$ to be orthogonal, either we are performing $$\langle (2,1),(0,0) \rangle$$ or it's true by definition.

It would be helpful if someone could explain this.

Regards

• What is $W$? Is it the set of a single vector $\lbrace (2,1)\rbrace$, is it the subspace spanned by $(2,1)$, or is it the inner product of something? Your notation is a bit confusing. Commented Oct 26, 2022 at 9:33
• @SomeCallMeTim W is a subspace Commented Oct 26, 2022 at 9:34
• okay, then by the usual definition of orthogonality, $W$ is not orthogonal. Indeed, it will contain $(2,1)$ and $(4,2)$, which are not orthogonal - their inner product will be $10$, which is not zero. What your source might have in mind is that you have an orthogonal basis for $W$, namely $\lbrace (2,1)\rbrace$. This is vacuously orthogonal, as you suggested. Commented Oct 26, 2022 at 9:39

If we define $$V = \{ v_{1}, v_{2}, \dots v_{n}\}$$ to be orthogonal if $$\left< v_{i}, v_{j} \right> = 0$$ when $$i \not = j$$, it is necessarily true that a one element set should be orthogonal (i.e $$n=1$$). There are a few ways to see this.
The first is by contrapositive: $$V$$ is not orthogonal if and only if there exists $$i,j$$ with $$i \not = j$$ and $$\left< v_{i}, v_{j} \right> \not = 0$$. However this cannot be, since $$i = j = 1$$ is the only possibility for $$i$$ and $$j$$. A second way is to see it as a vacuous truth. If we see the claim as "for all $$i,j$$ with $$i \not= j$$, $$\left< v_{i}, v_{j} \right> = 0$$", then essentially since there are no cases with $$i \not = j$$, we have that every case where $$i \not = j$$ also satisfies $$\left< v_{i}, v_{j} \right> = 0$$.
As pointed out in the comments, if $$W$$ denotes the subspace spanned by one non-zero vector, it is definitely not orthogonal. Taking the span necessarily means we have at least two non-zero vectors (really infinitely many) that are scalar multiples of each other, so their inner product is certainly non-zero. Generally this is what we see: given any set of nonzero vectors, orthogonal or not, taking the span is going to give something that is not orthogonal.