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
It's useful when it's applied to functions. It's just easier to introduce the concept of ____ spaces through vector spaces rather than functional spaces. At least that's what I've been told in class...
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!
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.
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
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! Yes expanding on what @OngoGablogian233 said, the vector space consists of all vectors should have the form [x, 2]. Adding [a1, 2] and [b1,2] gives [a1+b1,4] which is a different form than [x, 2]. We need to start with two vectors of form [x, 2] and end with a resulting vector of form [x, 2] in order to call the "set of all vectors of length 2 with 2 at the bottom" a vector space . Does that help?
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
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 ;---;
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?
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!
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.
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?
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.
