Dave you helped me all throughout high school, and now here you are helping me the night before my linear algebra exam while I'm in college studying for an aerospace engineering degree. You are the best !! :)
professor dave, you helped me through general chemistry 1 and 2 and organic chemistry and calculus 2 and physics mechanism, and now linear algebra. I am a heavy youtube learner, I never go to class I only learn by watching videos on youtube, and let me tell you, your videos are SERIOUSLY THE BEST in teaching the material, you are talented in teaching, you just make the material super easy for a 10 years old kid to understand. thank you.
wait you learned general chemistry 1 and 2 and organic chemistry and calculus 2 and physics mechanism and you are only 10 years old ????????????? How ?
@@jamestennant7789 i think he meant that prof dave is so good at explaining, a ten year old kid could understand. If I'm wrong, we'll probably hear about Hamid's breakthrough in science soon lol
Your explanation is so concise. Now I see that the vector space properties and behavior are the same as we learned in earlier classes, but I don't know what is wrong with college instructors. It is like they can't explain it straight forward. Thank you Prof. Dave.
this is lowkey the best vector space explanation. i mean everything was so nicely explained. this topic got over like a month ago in my college but i just couldnt comprehend its basics from any other youtube tutorial. so thnku thnku soooo much =) edit: i've literally wasted so much time watching super lengthy videos about vector space but this was so consice and simple that it really means a lot
Professor Dave, you have no idea how much you've been helping me. Even since i've started college, i've been overwhelmed with so many terms in physics and mathematics i didn't understand at first, but thanks to you, it's been much easier. So thank you
i've been following since your tagline was just about science! thanks for everything professor dave. i keep coming back cause you have by far the clearest and most concise explanations out there
Hey man, I just want to say, you explained this way better than my college professor. Dude is running his online class through and HTML page, not mentioning what kind of stuff he is putting on each quiz, and is insanely disorganized. His lectures also suck, are way too quiet, super disorganized, and take way too long. I’ve learned more about vector spaces in half of this video than two of his hour long lectures. I really appreciate it
Many moons ago (around 2002) I was studying Linear Algebra (Physics, at University). One of the reasons I left it was because I couldn't understand it at all. I saw your video, took a pencil, and I have understood it all. Thank you Professor Dave!
I dont understand why in example at 6:25 [a1, 2] is not a sample space. Its explained that its not a vector space because when you add it becomes [a1+b1,4] and the 4 is outside the initial [a1, 2]. but in the example at 4:48 when vector b is added to vector a it causes it to become [a1+b1,a2+b2,a3+b3] and the bs are adding to the a causing it to be outside the initial [a1,a2,a3].
Hi Dr Dave. This is really helpful and presented very well. You give theory and examples... Just superb delivery. Linear systems is kicking my butt. Thanks a ton.
Prof, how I wish I have you physically as my lecturer! You are an academic doctor who is EXCEPTIONALLY skilled in diagnosis of patients( your students) and provides drugs and injections 💉 ( the Fundamentals ) for healing( full understanding of concepts) Prof,please, I have serious challenges in REAL ANALYSIS and ABSTRACT ALGEBRA. Do you have dedicated videos on them or references that present the subject in a very rich manner? I give your work here 5 stars 🌟🌟🌟🌟🌟 sir
Do we all time need to multiply by scalar to see whether space is closed or not? for example: A space holds all vector which are, a [x, 0 -x]. now if we do scalar multiplication then we will get vectors like a[x, 0, -x] form. Again if we do addition then we will still get vectors of form a [x, 0 -x]. But if i multiply a with a, then i need to do [a (dot) transpose of a], then i can multiply. In such case i get resultant with different dimension. So can i say my space is vector space?
I cannot express just how much I love your videos, you single handedly managed to get me through first semester, and now you are saving my ass yet again ;---;
thank u so much sir for this video explanation but if we consider element ax+b as a polynomial belonging to vector space V and -ax+c also belong to V as its a linear polynomial but in this case, the closer property of addition will not be satisfied as we will get b+c which will not belong to V, so a set of liner polynomial s must not a vector space?? please sir can you please this doubt
I don't think that you mentioned that the set V must contain the zero "vector" to a vector space. So your last example where v = [a, 2]^T could never be a vector space because it doesn't contain the zero vector (ie it doesn't pass through the origin).
Also, with that 3rd property in mind, I think he is conflating the idea of vector spaces with the idea of subspaces. Subspaces must contain the zero vector, and have closure under scalar multiplication and addition. A vector space must satisfy the 8 properties he listed at the beginning of the video. A subspace is a vector space that satisfies the 3 additional aforementioned properties. All subspaces are vector spaces, but not all vector spaces are subspaces.
@@alexishemeon Hmm, I just want to make sure I understand. If both vector spaces AND subspaces must contain the zero vector, what is the difference between the two? Are the 8 properties he listed in the beginning the difference? In other words, the vector space requires "extra stuff" that the subspace does not? In other words, the 8 properties he listed in the beginning are also a requirement for a vector space in ADDITION to closure and zero vector inclusion (which are the only requirements for subspace)?
@@laulau4367 You shouldn't think of subspaces as needing "more" or "less" stuff than any other vector space. Instead, you should think of subspaces as answering the following question: If I have a known vector space V, and I have a _subset_ of vectors from V, when can I say that this subset is, in its own right, a vector space, using the same vector addition and scalar multiplication as V uses?" A lot of people do not emphasize the "same vector addition and scalar multiplication" part, but it's actually _super important_ here. So let's say you have a vector space V, and let's call your _subset_ W. In order to check that W is a vector space in its own right, we should check all of the axioms of a vector space. But because every vector in W is a vector in V and because W uses the same operations as V, a lot of the axioms are automatically true for W _because_ they are true for V. For example, one of the axioms of a vector space is to check that, for all vectors x and y in W, we need x+y = y+x. However, all vectors x and y in W are also vectors in V. And in V, we know that x+y = y+x. And since W is using the same vector addition as V, since we know x+y = y+x in V, we get that x+y = y+x in W too. A lot of the axioms of a vector space have this same sort of reasoning. They are automatically inherited by W since W is a subset of V and uses the same operations as V. The only axioms of a vector space which are _not_ automatically inherited by W are: closure under addition, closure under scalar multiplication, and the existence of the 0-vector. This is why the subspace test only requires you to check these three conditions. All the other conditions are automatically satisfied _because_ W is a subset of a known vector space and uses the same operations as that vector space.
False. It isn’t a vector of length 1 - it is a vector of dimension 1 and its magnitude is 5 ! -5 and 5 as vectors belonging to R^1 have the same magnitude - but assuming we use the Cartesian coordinate system - they are anti-parallel so the vectors -5 and 5 added give the 0 vector!
6:01 if a2=-a1, then the resulting linear equation would be a1x+b1+a2x+b2 --> a1x-a1x+b1+b2 = b1+b2 which is not contained in the set of linear polynomials. Does that mean it is not a vector space?
Great work, but don’t we have to examine that an object is non-empty to verify that it’s a vector space in addition to closures of addition and scalar multiplication?
The axioms requiring the existence of a 0-vector ensures that your set is nonempty. You may be confusing the concept of a general vector space with the subspace test. If you have a known vector space V and you have a subset W of V, how do you know whether or not W is a vector space (under the same vector addition and scalar multiplication as V) on its own? Since W shares the same operations as V, W inherits many of the axioms of a vector space from V being a vector space. The only ones which are not guaranteed are the two closure axioms and the existence of a 0-vector. Because 0v = 0-vector for all vectors v in V, it turns out that showing W has the 0-vector is equivalent to showing W is nonempty, provided you know W is closed under scalar multiplication. So you can replace "closed under addition and scalar multiplication and has the 0-vector" with "closed under addition and scalar multiplication and is nonempty". But as I pointed out, the above paragraph is the test of a subset of a known vector space being a subspace. If you have a set with an addition operation and a scalar multiplication operation, but if you don't know it's a subset of a known vector space with the same operations as that known vector space, then you have to check all of the vector space axioms.
Good sir? Professor?! Im just a "student" raising his hand (I won't take much of your time): I understand that I could recite to you - something like >1% - of the mathematics that you understand fluently. (I enrolled into an online college program many years ago and had to withdraw early, even directly after making honor roll.) As I understand all that I see written or shown or heard thrice, and after seeing how you hilariously handled the "flat heads," then seeing this video, I know I need to watch all of your channel in chronologic order. The things people could learn if they watched the two of us have a conversation is astronomical. You got the math part, I got the ethics part.
it's getting abstract. the laws on vector spaces 1:43 are not defined arbitrarily if some might think. They form an algebraic 'Field'. these rules are the same like when calculating with 'real numbers' (actually just what school math is about)
No, it does not form a field. Vectors have only one binary operation defined on them, that being addition (necessarily, you can of course define an inner product and other operations); fields require two. The vector set is instead an Abelian group. The product you see is between scalars and vectors. The scalars themselves must form a field. Together, the two sets form a vector space. This is distinct from a field.
Dave you helped me all throughout high school, and now here you are helping me the night before my linear algebra exam while I'm in college studying for an aerospace engineering degree. You are the best !! :)
bruuuh what a coincidence 'xD
@MIDDLE east where you studying at :p
Im at IST (Portugal)
same lol
what a coincidence i have a LinAlg exam tomorrow
This shows everyone searches for exams one day before....by the way I too have my sem end😂😂
professor dave, you helped me through general chemistry 1 and 2 and organic chemistry and calculus 2 and physics mechanism, and now linear algebra. I am a heavy youtube learner, I never go to class I only learn by watching videos on youtube, and let me tell you, your videos are SERIOUSLY THE BEST in teaching the material, you are talented in teaching, you just make the material super easy for a 10 years old kid to understand. thank you.
A 10-year-old kid wrote this and learned all that?
How's it going Hamid?
You must be 12 now
wait you learned general chemistry 1 and 2 and organic chemistry and calculus 2 and physics mechanism and you are only 10 years old ?????????????
How ?
@@jamestennant7789 i think he meant that prof dave is so good at explaining, a ten year old kid could understand. If I'm wrong, we'll probably hear about Hamid's breakthrough in science soon lol
Your explanation is so concise. Now I see that the vector space properties and behavior are the same as we learned in earlier classes, but I don't know what is wrong with college instructors. It is like they can't explain it straight forward. Thank you Prof. Dave.
Passed my math exam. Thnx from the Netherlands.
hi bro
I have been trying to learn this for an hour or two now and i m wondering why no one could explain this as simply as you. thank you so much
Well ...after attending so many college for hours ...I understood in 8 mins. thanks to you sir😁
Youre are singlehandedly carrying my liner algebra class in Uni, way better than my professor ever explained.
this is lowkey the best vector space explanation. i mean everything was so nicely explained. this topic got over like a month ago in my college but i just couldnt comprehend its basics from any other youtube tutorial. so thnku thnku soooo much =)
edit: i've literally wasted so much time watching super lengthy videos about vector space but this was so consice and simple that it really means a lot
Professor Dave, you have no idea how much you've been helping me. Even since i've started college, i've been overwhelmed with so many terms in physics and mathematics i didn't understand at first, but thanks to you, it's been much easier. So thank you
I still dont get the point of vector spaces
I think it’s to make “things” more general and abstract.
i've been following since your tagline was just about science! thanks for everything professor dave. i keep coming back cause you have by far the clearest and most concise explanations out there
Hey man, I just want to say, you explained this way better than my college professor. Dude is running his online class through and HTML page, not mentioning what kind of stuff he is putting on each quiz, and is insanely disorganized. His lectures also suck, are way too quiet, super disorganized, and take way too long. I’ve learned more about vector spaces in half of this video than two of his hour long lectures. I really appreciate it
Many moons ago (around 2002) I was studying Linear Algebra (Physics, at University). One of the reasons I left it was because I couldn't understand it at all. I saw your video, took a pencil, and I have understood it all. Thank you Professor Dave!
Concise yet Comprehensive! Perfect presentation!
After all the videos I watched, you were the one that helped me actually understand all of this. Thank you!
If you have any problem you can ask from me
Your explanation is better than MIT tutorials and also very concise.
Thank You Sir , I Had Hard Time Understanding Algebra , Thanks A Lot I Am Getting It , I Wish You Always Stay Happy And Healthy.
This dude is the best on youtube
I dont understand why in example at 6:25 [a1, 2] is not a sample space. Its explained that its not a vector space because when you add it becomes [a1+b1,4] and the 4 is outside the initial [a1, 2]. but in the example at 4:48 when vector b is added to vector a it causes it to become [a1+b1,a2+b2,a3+b3] and the bs are adding to the a causing it to be outside the initial [a1,a2,a3].
Same
I think in the space V, all vectors should have the form (a, 2). but when the vectors a+b are added the bottom row, becomes 4.
Explained it well for me in just 8 mins, thank you!
This lesson was so full and concise, it was the best! Thank you so much Professor Dave!!
Hi Dr Dave. This is really helpful and presented very well. You give theory and examples... Just superb delivery. Linear systems is kicking my butt. Thanks a ton.
Thanks so much! Your example at the end that showed when the vector is not closed under addition made everything click!
thank you for your simple explanation
Hey Prof, I really appreciate what you do and your videos are really helpful. Salute from Ethiopia 🇪🇹
Prof, how I wish I have you physically as my lecturer!
You are an academic doctor who is EXCEPTIONALLY skilled in diagnosis of patients( your students) and provides drugs and injections 💉 ( the Fundamentals ) for healing( full understanding of concepts)
Prof,please, I have serious challenges in REAL ANALYSIS and ABSTRACT ALGEBRA. Do you have dedicated videos on them or references that present the subject in a very rich manner?
I give your work here 5 stars 🌟🌟🌟🌟🌟 sir
After so much run and pain ..i got this .. the best video ever
My english isn't very good but your video is easy to understand. Thankyou prof
Thank you Prof Dave. Would appreciate more examples and bit longer videos
Thank you so much Professor Dave!!!! You're a life saver.
Do we all time need to multiply by scalar to see whether space is closed or not? for example: A space holds all vector which are, a [x, 0 -x]. now if we do scalar multiplication then we will get vectors like a[x, 0, -x] form. Again if we do addition then we will still get vectors of form a [x, 0 -x]. But if i multiply a with a, then i need to do [a (dot) transpose of a], then i can multiply. In such case i get resultant with different dimension. So can i say my space is vector space?
Awesome... You are really great...
Best wishes to you (by Md. Azmir Ibne Islam... From BRAC University Bangladesh)
Every subspace of R5 that contains a nonzero vector must contain a line. Is this statement true?
Very good presentation..and understandable😍👏
I love your videos. nice job they are extremely helpful
sir u didnt said why we use vector spaces . i know about vectors but in vector spaces im not clear.
Dude thank you for making this video. Really helped me understand vector spaces
professor dave sir you are the best among all
Crystal clear explanation! 😊
what a guy; came for terry crews, stayed for knowledge. actuarial science honours grad here
Excellent explanation
Beautiful explination!
you make life easier
Thank you sir 🙏🏼
Thanks, now I know what I'm getting myself into
Thank you for a very good tutorial.
why this channel is not growing fast!!!!! It is not fair...
I cannot express just how much I love your videos, you single handedly managed to get me through first semester, and now you are saving my ass yet again ;---;
great explanation
That was super helpful for a begineer
thank you, you just saved me from being lost!!🙏
Amazing explanations.. Great job
thanks prof
thank u so much sir for this video explanation but if we consider element ax+b as a polynomial belonging to vector space V and -ax+c also belong to V as its a linear polynomial but in this case, the closer property of addition will not be satisfied as we will get b+c which will not belong to V, so a set of liner polynomial s must not a vector space?? please sir can you please this doubt
Saving me before my final 🙏
who is here after not understanding GP sir video
This is just the kind of video I neded.Thx
well explained, thanks!
your teaching is understandable
Superb explanation❤
Its helps me
a lot sir😍😍
Thank you so much. This was really helpful.
Thanks alot you made very easy
I don't think that you mentioned that the set V must contain the zero "vector" to a vector space. So your last example where v = [a, 2]^T could never be a vector space because it doesn't contain the zero vector (ie it doesn't pass through the origin).
Also, with that 3rd property in mind, I think he is conflating the idea of vector spaces with the idea of subspaces. Subspaces must contain the zero vector, and have closure under scalar multiplication and addition. A vector space must satisfy the 8 properties he listed at the beginning of the video. A subspace is a vector space that satisfies the 3 additional aforementioned properties.
All subspaces are vector spaces, but not all vector spaces are subspaces.
@@alexishemeon Hmm, I just want to make sure I understand. If both vector spaces AND subspaces must contain the zero vector, what is the difference between the two? Are the 8 properties he listed in the beginning the difference? In other words, the vector space requires "extra stuff" that the subspace does not? In other words, the 8 properties he listed in the beginning are also a requirement for a vector space in ADDITION to closure and zero vector inclusion (which are the only requirements for subspace)?
@@laulau4367 You shouldn't think of subspaces as needing "more" or "less" stuff than any other vector space.
Instead, you should think of subspaces as answering the following question: If I have a known vector space V, and I have a _subset_ of vectors from V, when can I say that this subset is, in its own right, a vector space, using the same vector addition and scalar multiplication as V uses?"
A lot of people do not emphasize the "same vector addition and scalar multiplication" part, but it's actually _super important_ here.
So let's say you have a vector space V, and let's call your _subset_ W. In order to check that W is a vector space in its own right, we should check all of the axioms of a vector space. But because every vector in W is a vector in V and because W uses the same operations as V, a lot of the axioms are automatically true for W _because_ they are true for V.
For example, one of the axioms of a vector space is to check that, for all vectors x and y in W, we need x+y = y+x. However, all vectors x and y in W are also vectors in V. And in V, we know that x+y = y+x. And since W is using the same vector addition as V, since we know x+y = y+x in V, we get that x+y = y+x in W too. A lot of the axioms of a vector space have this same sort of reasoning. They are automatically inherited by W since W is a subset of V and uses the same operations as V.
The only axioms of a vector space which are _not_ automatically inherited by W are: closure under addition, closure under scalar multiplication, and the existence of the 0-vector. This is why the subspace test only requires you to check these three conditions. All the other conditions are automatically satisfied _because_ W is a subset of a known vector space and uses the same operations as that vector space.
@@MuffinsAPlentythanks man
this giving me modern algebra vibes
3:15 So the number 5 (and any real number) is both a scalar and a vector?
A vector of length 1
False. It isn’t a vector of length 1 - it is a vector of dimension 1 and its magnitude is 5 ! -5 and 5 as vectors belonging to R^1 have the same magnitude - but assuming we use the Cartesian coordinate system - they are anti-parallel so the vectors -5 and 5 added give the 0 vector!
amazing, spectacular, thank you
thanks dave
Best teacher ever
Sir please make video on Botany topics. Please sir because you are only source of learning easily.
why the heck i dont understand even a bit ... bruhhhhh its freaking driving my mind craaaaazaaayyyyyyyyyyyy
You'll get it bro. I believe in you.
do you understand it now buddy
@@nishadr.7637 I still dont get a shit about it 😂
So if the closure properties are met, can we assume all the other properties required for a vector space are met too?
THANK YOU
Professor Daves please make a video on youngs inequality,holders inequality and minkowski inequality?
6:01 if a2=-a1, then the resulting linear equation would be a1x+b1+a2x+b2 --> a1x-a1x+b1+b2 = b1+b2 which is not contained in the set of linear polynomials. Does that mean it is not a vector space?
You mistiped smth up there and it just means that a1=a2=0
Can you please do tutorials on abstract linear algebra?
Thanks!
thank you professor!
Thankyou Jesus
Thank you sir for this video.
But I still don't understand why number 1 in the comprehension is false.
Please can someone explain this 🙏🙏
I really loved intro song
Great work, but don’t we have to examine that an object is non-empty to verify that it’s a vector space in addition to closures of addition and scalar multiplication?
The axioms requiring the existence of a 0-vector ensures that your set is nonempty.
You may be confusing the concept of a general vector space with the subspace test. If you have a known vector space V and you have a subset W of V, how do you know whether or not W is a vector space (under the same vector addition and scalar multiplication as V) on its own?
Since W shares the same operations as V, W inherits many of the axioms of a vector space from V being a vector space. The only ones which are not guaranteed are the two closure axioms and the existence of a 0-vector. Because 0v = 0-vector for all vectors v in V, it turns out that showing W has the 0-vector is equivalent to showing W is nonempty, provided you know W is closed under scalar multiplication. So you can replace "closed under addition and scalar multiplication and has the 0-vector" with "closed under addition and scalar multiplication and is nonempty".
But as I pointed out, the above paragraph is the test of a subset of a known vector space being a subspace. If you have a set with an addition operation and a scalar multiplication operation, but if you don't know it's a subset of a known vector space with the same operations as that known vector space, then you have to check all of the vector space axioms.
Very precise and easy to understand explanation :) ! Thank you so much for this !
Dave I just love you man!
You're legend 🙏
what is the application of vector space in real life sir?
perfect
so we don't have to find all ten axioms to know if its a vector space????
Technically you do, it's just that checking all 10 properties takes a while. Some people skip some properties because of laziness.
thanksss
The best.
Good sir? Professor?! Im just a "student" raising his hand (I won't take much of your time): I understand that I could recite to you - something like >1% - of the mathematics that you understand fluently. (I enrolled into an online college program many years ago and had to withdraw early, even directly after making honor roll.) As I understand all that I see written or shown or heard thrice, and after seeing how you hilariously handled the "flat heads," then seeing this video, I know I need to watch all of your channel in chronologic order. The things people could learn if they watched the two of us have a conversation is astronomical. You got the math part, I got the ethics part.
It is help full!!!
Mathematicians with their “practical in theory” naming scheme
Vector spaces, aren’t really about vectors at all, just some fancy sets lol
Love you sir ❤❤❤❤❤
You are the best!!
Am watching this from Kyambogo university Uganda
it's getting abstract. the laws on vector spaces 1:43 are not defined arbitrarily if some might think. They form an algebraic 'Field'. these rules are the same like when calculating with 'real numbers' (actually just what school math is about)
No, it does not form a field. Vectors have only one binary operation defined on them, that being addition (necessarily, you can of course define an inner product and other operations); fields require two. The vector set is instead an Abelian group. The product you see is between scalars and vectors. The scalars themselves must form a field. Together, the two sets form a vector space. This is distinct from a field.
Explained so well even my grandma can understand it now🤣
goated, thank you
Your videos are good but the subtitles ruins it as i can't see anything
Then turn off subtitles.
7:31 why?