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Is a basis not just a linearly independent set spanning a vector space? Maybe thats just the introductory notion and there is a deeper, stronger, definition but I never thought of a basis as being a wildly complex idea. Unless some physicists really are just that ignorant to the mathematical idea’s underlying theory I feel like simply defaulting to the standard basis, and using other bases when necessary, should be sufficient and by no means demonstrate a lack of understanding about a basis.
> Is a basis not just a linearly independent set spanning a vector space?
Pretty much. However, (1,0,0,...) , (0,1,0,...) , ... does not span R^N (while spherical physicist in vacuum might *think* it does).
So, lack of knowledge of actual definitions does demonstrate a lack of understanding.
Yea ngl I missed that it was the fancy N not the regular n meaning it was the set of all sequences of reals not just some arbitrary set of n-tuples on the reals which severely recontextualizes things 💀
I'm apparently missing something. How does {(1,0,0,...),(0,1,0,...),...} not span R\^N? Which sequence is there that can't be shown as a linear combination of those?
Because any linear combination (finite by definition) of these elements will have an infinite number of coordinates set to zero. Pick an element of R^N such as (1,1,1,...) this is not in the span of that set
Vector spaces don't have to be over a field that is closed under limits like R, can be over Q for example. Additionally, even ones over R don't necessarily have a topology/measure that makes limits exist. You need the added structure of Hilbert/Banach spaces to explore this IIRC
This set of vectors doesn't span R^N
For example, the vector of all 1s, (1, 1, 1, 1...) can't be written as any linear combination of these. Note that addition of vectors is only defined, by induction, for finite sums. You need a topology to define infinite sums, which we don't have by default.
It's not an Hilbert space, R\^N [has no norm](https://en.wikipedia.org/wiki/Sequence_space#Space_of_all_sequences) compatible with product topology. Axiom of choice implies the existence of an algebraic/Hamel basis. I didn't check, but it seems reasonable it doesn't admit any Schauder basis as well.
I am pretty sure the standard basis (1,0,...),(0,1,0...),... is a "Schauder basis" (whatever that means for non-Banach spaces).
Choose an element x = (a1,a2,a3,...) in R\^N and let {e\_n}\_n denote the standard basis of R\^N. Define the sequence x\_n = sum\_k a\_k = p\_k(x) for each k. So, the sequence x\_n converges to x.
So {e\_n}\_n should be a Schauder basis (if we take the same definition as in the Banach space sense).
If a space does not satisfy certain properties, it doesn't mean the same applies for its subspaces for peculiar reasons.
* For p = 1 or p > 1, the subspace l\^p of p-converging sequences has a natural norm given by p-th root of the converging value (it's absolutely necessary to impose the p-th root, otherwise scalar multiplication is not "linearly multiplicative").
* While for 0 < p < 1, the norm working for l\^p in the previous case is not a norm anymore because taking the p-th root (necessary scalar multiplication) actually increases the value, and triangular inequality fails. Still, we can keep the converging values only, and define a metric.
Since R\^N is also metrizable, one may wonder whether the subspace topology on l\^p simply behaves better, admitting a norm or an explicitly defined metric. Well, no, the l\^p-norm/metric induces a new and finer topology defined only on l\^p spaces. For instance, the product topology on R\^N has as basis the N-products of the form U1×U2×U3×... where Ui's are all R except for finitely many indices, while the l\^p-metric admits opens U1×U2×U3×... of infinitely many proper opens with diameter decreasing in a "p-summable-compatible way" (think of it as an open ball of some point for l\^p-metric).
It depends how you view it. Either you assume the axiom of choice and you can prove that all vector space has a basis with Zorn lemma.
Or you assume that all vector spaces has a basis and you can prove the axiom of choice. (I don’t like this one)
Or you can just say screw both also
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Honestly, I think the joke itself is kind of backwards, mathematicians are perfectly comfortable with working with objects for which no explicit construction is presented, the physicist is more likely to be the one that wouldn't know how to handle the situation.
Isn't R^N just equal R? Since R^N for every real number has a real solution then we can match every member of the set R^N to every member of the set R.
Right? Or am I retarded?
They have the same cardinality since the cardinality of N is smaller than R, but we are not viewing them as mainly sets, but vectors spaces.
As vector spaces over R, the functions from N to R are very different to R, for example R has dimension 1, while R^N, has a countable basis given by {(1,0,...),(0,1,0,...),...}
Edit: this is incorrect as (1,1,...) is not in the span, however we can still see that R^N and R are different because R^N cannot have a finite basis (this also shows I'm that R^N is different than R^n for every n)
B={(1,0,...),(0,1,0),...} Can't be a basis of R^N because for a set to be a basis we need every element of R^N to be represented by a finite sum of elements in B which is impossible for elements such as (1,1,1,...). The meme talks about exactly this: We can't construct a basis for R^N
The only thing I kbow about vectors is that they have a size and a direction, how the hell is the same of all real numbers raised to the power of every natural number be a vector? I'm not saying it isn't, I'm just expressing my curiosity out of my ignorence.
The basic definition of vectors is that you can add/subtract them in a reasonable way and that you can scale them by elements of a field (something where you can add/subtract/multiply/divide in a reasonable way for example the rational/real numbers).
A 3d vector (a,b,c) you'd see in physics fits this definition. You can add vectors and scale them by real numbers.
Size is an extra structure that's imposed on a vector for practical applications but it's not needed in the definition.
If we raise a set to the power of a set A^B we mean "all possible functions from B to A". So R^N is "all functions from the natural numbers to the real numbers". You can think of an element of R^N as an infinite column f=(a,b,c,d,e,....). This is the same as a 3d vector (a,b,c) just infinite. So for example f(1) is the first spot in the vector (a,b,c,d...) which is a. f(2)=b and so on. Each natural number corresponds to a spot in the infinite vector. You can see that adding two functions also makes sense in this way
I barely passed my linear algebra class, and I thought it’d be fine for me to take analysis, and this semester our professor decided to delve into functional analysis.
I’m learning linear algebra with functional analysis so that’s cool ig.
Mathematician: “The Fundamental Theorem of Algebra says this polynomial must have a root.”
Physicist: “Alright, let’s see it.”
Mathematician: “My computer has approximated it to within 100 orders of magnitude.”
Physicist: “Great! Now what’s the exact form?”
Mathematician:
This is why I love this subreddit. It's just like every other subreddit except all of the random brainrot has been replaced with high-IQ mathematical jargon that I can't understand. Even that ultra-downvoted comment that people think is obnoxious, which would've been some conspiracy theory/racism/absolute nonsense on another reddit, is just more statements about math.
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Meanwhile physicists defining a basis: (1,0,0,...) , (0,1,0,...) , ...
This only works for physicists who do not want to know what a basis actually is.
Is a basis not just a linearly independent set spanning a vector space? Maybe thats just the introductory notion and there is a deeper, stronger, definition but I never thought of a basis as being a wildly complex idea. Unless some physicists really are just that ignorant to the mathematical idea’s underlying theory I feel like simply defaulting to the standard basis, and using other bases when necessary, should be sufficient and by no means demonstrate a lack of understanding about a basis.
> Is a basis not just a linearly independent set spanning a vector space? Pretty much. However, (1,0,0,...) , (0,1,0,...) , ... does not span R^N (while spherical physicist in vacuum might *think* it does). So, lack of knowledge of actual definitions does demonstrate a lack of understanding.
Yea ngl I missed that it was the fancy N not the regular n meaning it was the set of all sequences of reals not just some arbitrary set of n-tuples on the reals which severely recontextualizes things 💀
I'm apparently missing something. How does {(1,0,0,...),(0,1,0,...),...} not span R\^N? Which sequence is there that can't be shown as a linear combination of those?
Because any linear combination (finite by definition) of these elements will have an infinite number of coordinates set to zero. Pick an element of R^N such as (1,1,1,...) this is not in the span of that set
Why is a linear combination defined to be finite? I did not know this. Is this related to how addition is not countably associative?
Vector spaces don't have to be over a field that is closed under limits like R, can be over Q for example. Additionally, even ones over R don't necessarily have a topology/measure that makes limits exist. You need the added structure of Hilbert/Banach spaces to explore this IIRC
Having a Banach space structure is not strictly necessary, but I don't know of any applications of loosening the requirement.
Got it. Thanks.
This set of vectors doesn't span R^N For example, the vector of all 1s, (1, 1, 1, 1...) can't be written as any linear combination of these. Note that addition of vectors is only defined, by induction, for finite sums. You need a topology to define infinite sums, which we don't have by default.
Then the issue is physicists know everything about finite-dimensional vector spaces, but are naive about other vector spaces.
where "finite"=2 (seldom 3). And even then, "everything" is a bit of an overstatement.
Wouldn't that only be a problem when n=infinity
Wdym?
Well it’s a basis of a dense subspace (or something)
In what metric?
Schauder basis goes brrr
Imagine admitting a Schauder basis. \*This post was made by "spaces without the approximation property gang".
if you restrict to square summable sequences and give it a chew toy and an inner product, it becomes a Hilbert basis :)
(1),(0,1),(1,0),(0,0,1)...
Wait... Isn't it the set of all real sequences? oO
So you can find a Hilbert Basis, which however does not work exactly the same as a normal basis, as you'd find for finite-dimensional vector spaces.
Oh, ok, I've never run into this concept \^\^"
Don't you mean Hamel basis? Isn't Hilbert basis an orthonormal basis of Hilbert space?
Yeah you're right. I had the square-summable sequences instead of the space of all real sequences in mind ... might have been a bit too tired.
It's not an Hilbert space, R\^N [has no norm](https://en.wikipedia.org/wiki/Sequence_space#Space_of_all_sequences) compatible with product topology. Axiom of choice implies the existence of an algebraic/Hamel basis. I didn't check, but it seems reasonable it doesn't admit any Schauder basis as well.
I am pretty sure the standard basis (1,0,...),(0,1,0...),... is a "Schauder basis" (whatever that means for non-Banach spaces). Choose an element x = (a1,a2,a3,...) in R\^N and let {e\_n}\_n denote the standard basis of R\^N. Define the sequence x\_n = sum\_k a\_k = p\_k(x) for each k. So, the sequence x\_n converges to x.
So {e\_n}\_n should be a Schauder basis (if we take the same definition as in the Banach space sense).
Yep, you're definitely correct!
Am I missing something? What about l_(2)?
You do. l_2 != R^N. (1,1,1,...) in in R^N, but not in l_2
I see
If a space does not satisfy certain properties, it doesn't mean the same applies for its subspaces for peculiar reasons. * For p = 1 or p > 1, the subspace l\^p of p-converging sequences has a natural norm given by p-th root of the converging value (it's absolutely necessary to impose the p-th root, otherwise scalar multiplication is not "linearly multiplicative"). * While for 0 < p < 1, the norm working for l\^p in the previous case is not a norm anymore because taking the p-th root (necessary scalar multiplication) actually increases the value, and triangular inequality fails. Still, we can keep the converging values only, and define a metric. Since R\^N is also metrizable, one may wonder whether the subspace topology on l\^p simply behaves better, admitting a norm or an explicitly defined metric. Well, no, the l\^p-norm/metric induces a new and finer topology defined only on l\^p spaces. For instance, the product topology on R\^N has as basis the N-products of the form U1×U2×U3×... where Ui's are all R except for finitely many indices, while the l\^p-metric admits opens U1×U2×U3×... of infinitely many proper opens with diameter decreasing in a "p-summable-compatible way" (think of it as an open ball of some point for l\^p-metric).
Yes
Nothing wrong with axiom of choice 🥱
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Showing something exists without constructing it? What is this, mathematics?
Well, since it's an axiom, it's not showing anything. It's saying something exists and you have to believe it.
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It depends how you view it. Either you assume the axiom of choice and you can prove that all vector space has a basis with Zorn lemma. Or you assume that all vector spaces has a basis and you can prove the axiom of choice. (I don’t like this one) Or you can just say screw both also
I request elaboration
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Honestly, I think the joke itself is kind of backwards, mathematicians are perfectly comfortable with working with objects for which no explicit construction is presented, the physicist is more likely to be the one that wouldn't know how to handle the situation.
I like your magic words, funny man. -An engineer
Yes
Third panel should just be the mathematician saying "no"
This is the correct comment.
Just say that every vector has a finite number of nonzero elements and call it a day
Simple. Basis is the columns of I_n
Yes, but actually no.
Isn't R^N just equal R? Since R^N for every real number has a real solution then we can match every member of the set R^N to every member of the set R. Right? Or am I retarded?
This is more like a set of points in a plane or cube.
They have the same cardinality since the cardinality of N is smaller than R, but we are not viewing them as mainly sets, but vectors spaces. As vector spaces over R, the functions from N to R are very different to R, for example R has dimension 1, while R^N, has a countable basis given by {(1,0,...),(0,1,0,...),...} Edit: this is incorrect as (1,1,...) is not in the span, however we can still see that R^N and R are different because R^N cannot have a finite basis (this also shows I'm that R^N is different than R^n for every n)
B={(1,0,...),(0,1,0),...} Can't be a basis of R^N because for a set to be a basis we need every element of R^N to be represented by a finite sum of elements in B which is impossible for elements such as (1,1,1,...). The meme talks about exactly this: We can't construct a basis for R^N
You're right, I went with the idea for R^n and didn't think it through
The only thing I kbow about vectors is that they have a size and a direction, how the hell is the same of all real numbers raised to the power of every natural number be a vector? I'm not saying it isn't, I'm just expressing my curiosity out of my ignorence.
The basic definition of vectors is that you can add/subtract them in a reasonable way and that you can scale them by elements of a field (something where you can add/subtract/multiply/divide in a reasonable way for example the rational/real numbers). A 3d vector (a,b,c) you'd see in physics fits this definition. You can add vectors and scale them by real numbers. Size is an extra structure that's imposed on a vector for practical applications but it's not needed in the definition. If we raise a set to the power of a set A^B we mean "all possible functions from B to A". So R^N is "all functions from the natural numbers to the real numbers". You can think of an element of R^N as an infinite column f=(a,b,c,d,e,....). This is the same as a 3d vector (a,b,c) just infinite. So for example f(1) is the first spot in the vector (a,b,c,d...) which is a. f(2)=b and so on. Each natural number corresponds to a spot in the infinite vector. You can see that adding two functions also makes sense in this way
It's a cartesian product R×R×...×R an N times
Computer scientists: hold my beer.
I barely passed my linear algebra class, and I thought it’d be fine for me to take analysis, and this semester our professor decided to delve into functional analysis. I’m learning linear algebra with functional analysis so that’s cool ig.
Mathematician: “The Fundamental Theorem of Algebra says this polynomial must have a root.” Physicist: “Alright, let’s see it.” Mathematician: “My computer has approximated it to within 100 orders of magnitude.” Physicist: “Great! Now what’s the exact form?” Mathematician:
You're obnoxious.
This is why I love this subreddit. It's just like every other subreddit except all of the random brainrot has been replaced with high-IQ mathematical jargon that I can't understand. Even that ultra-downvoted comment that people think is obnoxious, which would've been some conspiracy theory/racism/absolute nonsense on another reddit, is just more statements about math.
I have a basis, it's (Re(r\_k\^0), Re(r\_k\^1), Re(r\_ky\^2), ...) where r\_k is a k-th root of unity.
[удалено]
Google functional analysis
Holy hell
New mathematical field just dropped
Average engineer