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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

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  • $\begingroup$ 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

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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.

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