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.


  • $\begingroup$ 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. $\endgroup$ Commented Oct 26, 2022 at 9:33
  • $\begingroup$ @SomeCallMeTim W is a subspace $\endgroup$ Commented Oct 26, 2022 at 9:34
  • $\begingroup$ 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. $\endgroup$ Commented Oct 26, 2022 at 9:39

1 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .