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how do you apply linear/abstract algebra regarding direct sums to boolean logic and xnor gates? I know what the tables look like but how would it be worked out mathematically?
Hello Sir, i have learnt about Direct Product also in abstract algebra defined over sets,there they said they need not be abelian groups,but from the book( serge lang undergraduate algebra) i Learnt about this direct sum Tells the two groups must be abelian....why so?please explain
@@paulfoss5385 sir, You seem to have an active channel if you know anything about these things, including mine and your question both,do make a video. Thank you
@@mathematicsgurucool7949 I do want a video that goes over the difference between the direct product and sum in different areas of mathematics and how these differences follow from the definitions, but I'm not very skilled with linear algebra, which is why I deleted my comment in fear I could end up looking foolish, especially since one of the things I brought up in that comment was how there were many resources online that state that infinite direct sums of vector spaces must have finitely many non zero terms, but infinite direct products have no such restriction, but those online resources never explain how that follows from the definition. But then I continued watching this video and it became clear why, at least for a finite dimensional space, that should be the case for the direct sum. But I'm still unsure about the definition of a direct product of vector spaces and why that differs. And anyway that still wouldn't negate the usefulness of having a video dedicated to their difference. Out of fear of embarrassment I ended up with a far longer and more embarrassing comment. 😂 Anyway if you're looking at my channel, check out "Unusual Factoring" it' my best video and only 20 sec.
Yea,so after a long time i got that answer.....,the group need to be abelian in direct sum beacuse then only the function will we we'll defined that defined g:G-->H×K,the conditions for g to be direct sum(sometime called internal direct product) of must be direct product of H and K,secondly H and K must be disjoint and third is the abelian property for xy=yx,x€H, and y€K,....now if i define the function g(g_1)=(h_1,k_1),if g_1=h_1*k_1 and it is walwys the case their existence is guaranteed beacuse G is Direct Product of those H and K,so g(g_1*g_2)=g(h_1*g_1*h_2*g_2)--(i),now if they are abelian then only it is homomorphism type,g(g_1)*g(g_2)=(h_1,k_1)*(h_2,k_2)=g(h_1h_2g_1g_2),now this last term is equal to that (i) if and only if h_2*g_1=g_1*h_2 and thus that abelian conditions on that group was needed..... I had this doubt it got cleared(due to Judson book on abstract algebra) and hence i answered it myself very weird and awkward😅😅
Sir, your looks are absolutely like a rapper......you are totally opposite of the usual professors of Oxford University in terms of appearance that comes in my mind......You totally have proven that..... don't judge a book by it's cover....😊😊😊
Here's how I understand this, pls correct me if I'm wrong. There are two types of direct sums (assuming done over finitely many objects): 1. Internal direct sum. This is the idea of, in one single big space V (usually a vector space), splitting V into disjoint subsets s.t. each element from V can be uniquely rewritten as a sum of many components, one from each subset. Like "each sniper combat team consist of a shooter and a spotter". Here x+y=v makes sense since both x,y live in a common space where "+" is defined. In this case you can roughly say "the direct sum of two v-spaces is again a v-space" and the dimensions are added together. 2. External direct sum. This is the same as direct product, cartesian product or tensor product, namely, the ideal of gluing objects that don't have anything to do with each other together and an element in there is an "ordered pair". If each object has its own operation, then the operation of the elements in the sum is done component-wise or independently. Everything stays separate. Take v from V, w from W, you get (v,w) in V x W. You don't do "v+w" since this is undefined and it's not equal to anything. And tensor product is simply adding one more meaning to (v,w), by the universal property: uniquely rewrite it as v \otimes w and define this as a "multilinear map". In this case, "the direct sum of groups/rings/fields/v-spaces" may no longer be another group/ring/field/v-space. You need to carefully test the definitions. And the dimensions are multiplied (because they are independent and it's like counting handshakes or all possible combinations).
"the sum" U+W of two subspaces is the set of sums of arbitrary vectors from U and V. in general if a vector v can be expressed as u+w where u is in U and w in W, then this representation need not be unique, but if it is for all v then U+W is a *direct sum* and you can write U oplus W. the uniqueness condition is helpful in other parts of linear algebra for example if V is a direct sum of U and W then the dimension (the smallest number of vectors in V needed to identify all points in space) is simply the sum of the dimensions of U and W im not sure why its specifically called a direct sum but this is enough to explain that "the sum" and "direct sum" are two different concepts
You can have vector sum only when there is a direct physical relationship between 2 different vector. In order to see it more clearly, we can use physics to explain what actually a vector is. It's mathematical representation of a motion, and a motion obviously has a speed, and a direction. Hence a vector = speed + direction . The speed represent the applied force acting on the moving object. When two vectors intersect, the outcome will be the interactions of two different kinetic forces which are acting on the two moving objects, resulting in a 3rd force which deflects the motion of the two objects in two other directions, and the magnitudes of the speeds of the 2 objects and their two other new directions are the very direct sum of space vectors because each vector alone always is the sum of 3 different vectors pointing toward 3 different directions in the 3-D space
A physics vector is a specific example of a vector. There are other vector spaces besides R^n. The space of all polynomials is a vector space which is not inherently physics related
if one of the 3 space vectors DNE or is non-existent, then you won't have any 3-D space, instead of that, you will have only a flat surface in human concepts of mathematics, although in objective reality, there is not any actual flat surface, but there are only and always 3-d spaces. Even a single point alone is also a representation of a 3-d space, no matter how extremely small it is because such small points as subatomic particles invisible to human eyes are merely a sort of down-scaled values of visible objects, in terms of the sizes, in the quantum world
The space vector( 1, 1, 1) is called as unit space vector, because other space vectors are merely scalar values of that basic unit space vector, by scaling up to larger or down to smaller than 1. But again, such human invented numbers alone have no actual meaning nor value without any specific and tangible physical world because 1 can be 1 cm or 1 m or 1 km or whatever you want
It is nonsensical or obscure to use mathematical operations to explain why the intersection of two vector must be zero because as mentioned earlier, 0 never exists anywhere in the universe. The only correct way to explain about the intersection part of 2 different vectors is that it is the starting point of the 3rd and resultant vector from the two intersecting vector. It means that the intersection part of the different vector can never be nought or 0, but always have a certain value in a volume space or a 3-D space. The symbolic mathematical value of 0 here can only represents either the equilibrium state of a point in space or a stationary object when all the applied forces acting on it are equal, like in your example where x' = x , y'' = y , and the point or the stationary intersection where the object located can never ever actually be 0 as how your mathematical expressions nonsensically says. Human defective concepts of mathematics are the very problems causing human fundamental misunderstandings of natural phenomena, and paradoxical theories in all branches of physics and human current development toward self-destruction
Mathematical representations alone without any physical world or any form of matter are both vague and useless, and many times misleading with manipulations of the flaws and defective holes of the concepts of mathematics invented by human limited knowledge. One of the most typical examples is the empty set, which never ever exists anywhere in the universe other than human wild imagination because zero is not any actual value, nor can it represent any form of matter at all, like human -invented concept of infinity
hey tom, could you try the putnam maths exam? it is one of Americas hardest exams and last i checked only 4 people have ever got a perfect score. It would be really interesting to see how you do on it. Thank you!
Considering he's a professor it would be trivial for him. Putnam is for undergrads and is mostly really challenging analysis, combinatorics and linear algebra.
Check out ProPrep with a 30-day free trial to see how it can help you to improve your performance in STEM-based subjects: www.proprep.uk/info/TOM-Crawford
Great information and learning as always Tom. Math rocks!
Thanks man. You just got another subscriber.
GED is way different than GCSE in United Kingdom!@samadesh5678
Could have done with this several years ago in my math undergrad
Great introduction :)
how do you apply linear/abstract algebra regarding direct sums to boolean logic and xnor gates? I know what the tables look like but how would it be worked out mathematically?
Very nice.
@Tom Rocks Math So, Do you teach GED?
Hello Sir,
i have learnt about Direct Product also in abstract algebra defined over sets,there they said they need not be abelian groups,but from the book( serge lang undergraduate algebra) i Learnt about this direct sum Tells the two groups must be abelian....why so?please explain
@@paulfoss5385 sir,
You seem to have an active channel if you know anything about these things, including mine and your question both,do make a video.
Thank you
@@mathematicsgurucool7949 Whoops, sorry, I deleted my comment out of feeling shy about things. I really shouldn't have.
@@mathematicsgurucool7949 I do want a video that goes over the difference between the direct product and sum in different areas of mathematics and how these differences follow from the definitions, but I'm not very skilled with linear algebra, which is why I deleted my comment in fear I could end up looking foolish, especially since one of the things I brought up in that comment was how there were many resources online that state that infinite direct sums of vector spaces must have finitely many non zero terms, but infinite direct products have no such restriction, but those online resources never explain how that follows from the definition. But then I continued watching this video and it became clear why, at least for a finite dimensional space, that should be the case for the direct sum. But I'm still unsure about the definition of a direct product of vector spaces and why that differs. And anyway that still wouldn't negate the usefulness of having a video dedicated to their difference.
Out of fear of embarrassment I ended up with a far longer and more embarrassing comment. 😂
Anyway if you're looking at my channel, check out "Unusual Factoring" it' my best video and only 20 sec.
@@paulfoss5385 sure sure,i see 🙈
Yea,so after a long time i got that answer.....,the group need to be abelian in direct sum beacuse then only the function will we we'll defined that defined g:G-->H×K,the conditions for g to be direct sum(sometime called internal direct product) of must be direct product of H and K,secondly H and K must be disjoint and third is the abelian property for xy=yx,x€H, and y€K,....now if i define the function g(g_1)=(h_1,k_1),if g_1=h_1*k_1 and it is walwys the case their existence is guaranteed beacuse G is Direct Product of those H and K,so g(g_1*g_2)=g(h_1*g_1*h_2*g_2)--(i),now if they are abelian then only it is homomorphism type,g(g_1)*g(g_2)=(h_1,k_1)*(h_2,k_2)=g(h_1h_2g_1g_2),now this last term is equal to that (i) if and only if h_2*g_1=g_1*h_2 and thus that abelian conditions on that group was needed.....
I had this doubt it got cleared(due to Judson book on abstract algebra) and hence i answered it myself very weird and awkward😅😅
cool ass teacher thx for the knowledge
Sir, your looks are absolutely like a rapper......you are totally opposite of the usual professors of Oxford University in terms of appearance that comes in my mind......You totally have proven that..... don't judge a book by it's cover....😊😊😊
Here's how I understand this, pls correct me if I'm wrong. There are two types of direct sums (assuming done over finitely many objects):
1. Internal direct sum. This is the idea of, in one single big space V (usually a vector space), splitting V into disjoint subsets s.t. each element from V can be uniquely rewritten as a sum of many components, one from each subset. Like "each sniper combat team consist of a shooter and a spotter". Here x+y=v makes sense since both x,y live in a common space where "+" is defined. In this case you can roughly say "the direct sum of two v-spaces is again a v-space" and the dimensions are added together.
2. External direct sum. This is the same as direct product, cartesian product or tensor product, namely, the ideal of gluing objects that don't have anything to do with each other together and an element in there is an "ordered pair". If each object has its own operation, then the operation of the elements in the sum is done component-wise or independently. Everything stays separate. Take v from V, w from W, you get (v,w) in V x W. You don't do "v+w" since this is undefined and it's not equal to anything. And tensor product is simply adding one more meaning to (v,w), by the universal property: uniquely rewrite it as v \otimes w and define this as a "multilinear map". In this case, "the direct sum of groups/rings/fields/v-spaces" may no longer be another group/ring/field/v-space. You need to carefully test the definitions. And the dimensions are multiplied (because they are independent and it's like counting handshakes or all possible combinations).
Yes, this is correct.
What is 'direct' about the direct sum? Why cannot we simply call it the sum?
"the sum" U+W of two subspaces is the set of sums of arbitrary vectors from U and V. in general if a vector v can be expressed as u+w where u is in U and w in W, then this representation need not be unique, but if it is for all v then U+W is a *direct sum* and you can write U oplus W. the uniqueness condition is helpful in other parts of linear algebra for example if V is a direct sum of U and W then the dimension (the smallest number of vectors in V needed to identify all points in space) is simply the sum of the dimensions of U and W
im not sure why its specifically called a direct sum but this is enough to explain that "the sum" and "direct sum" are two different concepts
😮
That hair is a direct sum of a buzz cut and justin bieber hair
😳
You can have vector sum only when there is a direct physical relationship between 2 different vector. In order to see it more clearly, we can use physics to explain what actually a vector is. It's mathematical representation of a motion, and a motion obviously has a speed, and a direction. Hence a vector = speed + direction . The speed represent the applied force acting on the moving object. When two vectors intersect, the outcome will be the interactions of two different kinetic forces which are acting on the two moving objects, resulting in a 3rd force which deflects the motion of the two objects in two other directions, and the magnitudes of the speeds of the 2 objects and their two other new directions are the very direct sum of space vectors
because each vector alone always is the sum of 3 different vectors pointing toward 3 different directions in the 3-D space
A physics vector is a specific example of a vector. There are other vector spaces besides R^n. The space of all polynomials is a vector space which is not inherently physics related
youre a term too late!!!! this wouldve helped me so much😭😭😭
if one of the 3 space vectors DNE or is non-existent, then you won't have any 3-D space, instead of that, you will have only a flat surface in human concepts of mathematics, although in objective reality, there is not any actual flat surface, but there are only and always 3-d spaces. Even a single point alone is also a representation of a 3-d space, no matter how extremely small it is because such small points as subatomic particles invisible to human eyes are merely a sort of down-scaled values of visible objects, in terms of the sizes, in the quantum world
The space vector( 1, 1, 1) is called as unit space vector, because other space vectors are merely scalar values of that basic unit space vector, by scaling up to larger or down to smaller than 1. But again, such human invented numbers alone have no actual meaning nor value without any specific and tangible physical world because 1 can be 1 cm or 1 m or 1 km or whatever you want
(2,1,1) is not a scalar value of (1,1,1), this is simply incorrect
It is nonsensical or obscure to use mathematical operations to explain why the intersection of two vector must be zero because as mentioned earlier, 0 never exists anywhere in the universe. The only correct way to explain about the intersection part of 2 different vectors is that it is the starting point of the 3rd and resultant vector from the two intersecting vector. It means that the intersection part of the different vector can never be nought or 0, but always have a certain value in a volume space or a 3-D space. The symbolic mathematical value of 0 here can only represents either the equilibrium state of a point in space or a stationary object when all the applied forces acting on it are equal, like in your example where x' = x , y'' = y , and the point or the stationary intersection where the object located can never ever actually be 0 as how your mathematical expressions nonsensically says.
Human defective concepts of mathematics are the very problems causing human fundamental misunderstandings of natural phenomena, and paradoxical theories in all branches of physics and human current development toward self-destruction
Cool story bro.
yap yap yap
this is math man not yappalogy
Your mathematical language is vague because "space" can be interpreted as a point, a 2 dimensional one, such as plane or flat surface or a 3-D space
He's obviously talking about a vector space that is rigorously defined.
Mathematical representations alone without any physical world or any form of matter are both vague and useless, and many times misleading with manipulations of the flaws and defective holes of the concepts of mathematics invented by human limited knowledge. One of the most typical examples is the empty set, which never ever exists anywhere in the universe other than human wild imagination because zero is not any actual value, nor can it represent any form of matter at all, like human -invented concept of infinity
hey tom, could you try the putnam maths exam? it is one of Americas hardest exams and last i checked only 4 people have ever got a perfect score. It would be really interesting to see how you do on it. Thank you!
Considering he's a professor it would be trivial for him. Putnam is for undergrads and is mostly really challenging analysis, combinatorics and linear algebra.