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For some reason my book distinguishes the two names.

If a set is an orthogonal set, doesn't that make it immediately a basis for some subspace $W$ since all the vectors in the orthogonal set are linearly independent anyways? So why do we have two different words for the same thing?

Dylan Moreland
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mim
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  • In the text from which I'm teaching this quarter (I think the author is Lay), an orthogonal set is allowed to contain the zero vector, which obviously precludes the set from being a basis for anything. An orthogonal set consisting of non-zero vectors is an orthogonal basis for its span, I agree. – Dylan Moreland Feb 22 '12 at 22:43
  • Are orthogonal sets always defined as subsets of some fixed vector space $V$? Because then the word "basis" would imply (at least to me) that they are a basis for $V$, not just for some subspace. – mdp Feb 22 '12 at 22:45
  • @Dylan, yeah that's the book I am using! – mim Feb 22 '12 at 22:46
  • Can you please quote your book's definition of "orthogonal set"? –  Feb 22 '12 at 23:21
  • A set of vectors $\left {u_1,...,u_p} \right .$ in $\mathbb{R}^n$ is said to be orthogonal set if each pair of distinct vectors from the set is orthogonal, that is, if $u_i \cdot u_j = 0$ whenever i is not equal to j – mim Feb 23 '12 at 00:06
  • @mim: By convention, if we say "basis" without specifying basis for what, we mean the basis of the whole space of discourse. – Arturo Magidin Feb 23 '12 at 03:44

4 Answers4

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When you say orthogonal basis you mean that the set is a basis for the whole given space. Every orthogonal set is a basis for some subset of the space, but not necessarily for the whole space.

Dennis Gulko
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  • How could that be? Could you give me an example? I can't visualize how that is possible – mim Feb 22 '12 at 22:51
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    Take your favorite orthogonal basis for your favorite vector space. Now, throw away one of the vectors in the basis. What's left is still a set of orthogonal vectors, but it's no longer a basis for your vector space. But maybe your definition of orthogonal set has more to it than just "set of orthogonal vectors"? – Gerry Myerson Feb 22 '12 at 23:15
  • OKay, I took $\mathbb{R}^2$ and $\hat{i}$ and $\hat{j}$, if I throw one away, then it wouldn't be a basis and one vector can't be an orthogonal set now can it? So I don't understand your argument – mim Feb 22 '12 at 23:19
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    Actually, one vector is an orthogonal set. – Thomas Andrews Feb 22 '12 at 23:38
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    ${(1,0,0) , (0,1,0)}$ is an orthogonal set of vectors in $\mathbb{R}^3$, but it is not an orthogonal basis of $\mathbb{R}^3$. – JeffE Feb 22 '12 at 23:49
  • Actually, the empty set is also an orthogonal set. – JeffE Feb 22 '12 at 23:49
  • @Jeff, can I generalize like for an orthogonal set $S = {u_1, ..., u_p}$ for $\mathbb{R}^n$ and it is a basis iff n = p, but it is not a basis if 1 < p < n? – mim Feb 23 '12 at 00:02
  • Except zero vector. It isn't a basis for any subset. And $u_i$ shouldn't be zero for all $i$. – sas Feb 23 '12 at 00:03
  • But am I right though? Because my book justifies by stating it is a basis because the set contains linearly independent vectors – mim Feb 23 '12 at 00:13
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    If you have an orthogonal set in ${\bf R}^n$, then it's not a basis for ${\bf R}^n$ if it has fewer than $n$ members, and it's not a basis for ${\bf R}^n$ if it contains the zero vector, and it is a basis for ${\bf R}^n$ if it has $n$ members and doesn't contain the zero vector. – Gerry Myerson Feb 23 '12 at 11:35
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The reason for the different terms is the same as the reason for the different terms "linearly independent set" and "basis".

Every linearly independent set is a basis for the subspace it spans. But when working in a larger space "basis" means "maximal linearly independent set" (not just spanning a subspace but spanning the whole thing).

An orthogonal set (without the zero vector) is automatically linearly independent. So we have "orthogonal sets" and then maximal ones are "orthogonal bases".

Note: In the end we're essentially just tacking on the adjective "orthogonal". We don't keep the words "linearly independent" in "orthogonal linearly independent set" because they're redundant.

Bill Cook
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  1. These two concepts are totally different.
  2. For "orthogonal set"$M$, we only have Bessel's inequality.
  3. But, if $M$ is orthogonal basis, then we get the Parseval's theorem. The key point is the completeness of this set M in your space. For example, in finite dimensional space $\mathcal{R}^3$, $\{i,j\}$ is an orthnormal set, but not an orthonormal basis. A common orthonormal basis is $\{i,j,k\}$.
Yong Yang
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If a set is an orthogonal set that means that all the distinct pairs of vectors in the set are orthogonal to each other. Since the zero vector is orthogonal to every vector, the zero vector could be included in this orthogonal set. In this case, if the zero vector is included in the set of vectors, the set is not linearly independent (thus this set cannot be a basis for some subspace since the vectors in the set are not linearly independent).

So a set being a basis for some subspace and a set being an orthogonal set are two different things. Now, it is possible for an orthogonal set (that does not include the zero vector) to be a basis for the subspace that it spans (and in this case we call that set of vectors an orthogonal basis).

user669076
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