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I am pretty confused about the relation and the difference between these two concepts. At first glance, when I hear: Orthonormal basis I instantly relates it to the standard basis - but I'm not sure about this relation.

Firstly, if these two concepts were the same - why have they got different names? Secondly, I have a sneaking suspicion that these are fundamentally different concepts - where one is mentioned when we concern about an inner product space and the other one (standard basis) is relevant when we represent column vectors over a field.

Nevertheless, I have no satisfying explanation to myself about what is the intrinsic difference between the both concepts.

So, can someone clarify this intrinsic difference between the concepts?

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  • $\begingroup$ Note: the standard basis for $\mathbb R^n$ is an orthonormal basis; you are correct that orthonormal basis is relevant for an inner product space $\endgroup$ – J. W. Tanner Jun 5 at 13:14
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    $\begingroup$ When working in vector spaces with inner products, the standard basis is one example of an orthonormal basis, but not the only one. These 2 vectors are an orthonormal basis of $\mathbb R^2$, but not the standard basis: $\{(-1,0), (0,-1)\}$. $\endgroup$ – kimchi lover Jun 5 at 13:16
  • $\begingroup$ @J.W.Tanner So it's pretty not clear for me. Is it a standard basis for $\mathbb R^n$ over which inner product? $\endgroup$ – HelpMe Jun 5 at 13:16
  • $\begingroup$ the inner product I'm talking about is the dot product of vectors in $\Bbb R^n$ $\endgroup$ – J. W. Tanner Jun 5 at 13:18
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Let's concentrate on the reals, in fact on $\Bbb R^3$. The vectors are column vectors containing two real numbers. Because $0$ and $1$ are special real numbers, it turns out to be really nice to work with vectors that are mostly zeroes. So $$ e_1 = \pmatrix{1\\0\\0}, e_2 = \pmatrix{0\\1\\0}, e_3 = \pmatrix{0\\0\\1}, $$ which turns out to be a basis for 3-space, are a really nice set. They come up a lot, so they get a name: "The standard basis". This generalizes to $\Bbb R^n$, and I'm pretty sure you understand the pattern.

As it happens, when we use the "standard inner product" on $\Bbb R^n$, these vectors turn out to all have length one, and be mutually perpendicular. Those two properties also come up a lot, so we give them a name: we say the basis is an "orthonormal" basis.

So at this point, you see that the standard basis, with respect to the standard inner product, is in fact an orthonormal basis.

But not every orthonormal basis is the standard basis (even using the standard inner product). For instance, in $\Bbb R^2$, for any value you pick for $t$, the vectors $$ v_1 = \pmatrix{\cos t \\ \sin t} , v_2 = \pmatrix{\sin t\\ \cos t} $$ are and orthonormal basis as well. (They're just the standard basis rotated counterclockwise by an angle $t$.)

The phrase "orthonormal basis" is always qualified with "with respect to ..." or "under ...", and then an inner product gets named. Well ... not always. Sometimes we're in the middle of talking about some inner product, and it's implicit. But a basis that's orthonormal with respect to one inner product may not be orthonormal with respect to another. Consider, on $\Bbb R^2$, the inner product defined by $$ \langle \pmatrix{a\\b} , \pmatrix{c\\d} \rangle = ac + 2bd. $$ Under this inner product, the vector $e_2$ has length $4$, so $\{e_1, e_2 \}$ is not an orthonomal basis with respect to this (peculiar) inner product.

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  • $\begingroup$ This is a really good answer. So I learned two major facts: first, the standard basis is always an orthonormal basis in respect to the standard inner product. Secondly, the standard basis is not an orthonormal basis in respect to the all possible inner products we have. Right? Thank you! $\endgroup$ – HelpMe Jun 5 at 13:26
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    $\begingroup$ Both those things are correct, yes. In fact, the standard basis is an orthonormal basis only with respect to the standard inner product. $\endgroup$ – John Hughes Jun 5 at 13:30
  • $\begingroup$ Good to know. Thank you! $\endgroup$ – HelpMe Jun 5 at 13:31
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    $\begingroup$ At the very least, I'd suggest writing "As you study more general vector spaces, such as those over finite fields, you'll learn that things are more subtle in those situations --- there may not even be a 'standard inner product'!" That'd be more likely to help OP rather than to confuse with terms that OP almost certainly hasn't encountered (like "formally real fields"). $\endgroup$ – John Hughes Jun 5 at 13:45
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    $\begingroup$ @J.W.Tanner: thanks, fixed. $\endgroup$ – John Hughes Jun 6 at 0:35
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They're not the same. The standard basis of $\mathbb R^n$ is$$\bigl\{(1,0,0,\ldots,0,0),(0,1,0,\ldots,0,0),\ldots,(0,0,0,\ldots,1,0),(0,0,0,\ldots,0,1)\bigr\}.$$It turns out that it as an orthonormal basis, but there are others. For instance, $\left\{\left(\frac35,\frac45\right),\left(-\frac45,\frac35\right)\right\}$ is also an orthonormal basis of $\mathbb R^2$.

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  • $\begingroup$ not only orthogonal, but orthonormal $\endgroup$ – J. W. Tanner Jun 5 at 14:36
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    $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jun 5 at 14:37

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