Why is a set containing a single vector orthogonal?

Question

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

Answer 1

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.