I had binge-watched your linear algebra videos and found it extremely helpful. I must say your teaching skills are so clear, simple, and very easy to understand. I wish I had such a teacher liked you back in my college years. Thank you very much for spending your time to prepare these videos!
Hi, could you please continue the series? Its so helpful for all students doing first year maths for engineering. they are so concise and cover all the important points. pleeeaaaaase!
So sad 😢to see the end of this amazing journey. I don't know how busy you are. I would really say if you have time and good health then please make video on all topics of abstract algebra. I'm not only talking about me. It'll be beneficial for whole humanity bec no one teach like you and student don't understand it and not able to apply this for betterment. And if you have any paid lecture then please let me know. I'm in love ❤ with Algebra and what to understand it more and more. So that one day Algebra starts loving me.
@@TheLazyEngineer how we can see a Matrix as linear mapping Bec for mapping we required two vector space or even one in case of self mapping, we have some algebraic expression and domain, codomain and range. But how we can see a Matrix in the same way. I means what would be a domain, and codomain. What would be the mapping rule which will take doman as input and product element from codomain as output/range.
@@alokraj8972 Consider 2 vector spaces, A and B. Let {a1, a2, ... aN} be a basis for A, and {b1, b2, ...bM} a basis for B. Now, let x be a vector in A. Then there exists some coefficients, {c1, c2, ...cN} such that x = c1 a1 + c2 a2 + ... + cN aN. Now, let T be a linear transformation from A to B. By the definition of linearity, we have: Tx = T(c1 a1 + c2 a2 + ... + cN aN) = c1 (T a1) + c2 (T a2) + ... + cN (T aN), where (T ai) = d1 b1 + d2 b2 + ... + bM dM, for some ceofficeints dM. If you think about it, you'll realize that you can write T as a matrix. Assuming the bases for A and B are the canoncial euclidean basis where ai = bi = [0,0,...,1,...0] with a 1 in the i'th position, then the (i,j)th elements of T are e_i (T e_j). When you think about linear transformations in the abstract sense as we have here, then you'll realize that the idea of a matrix and the rules about how we multiply them follow directly from these ideas.
Tbh it is helpful for engineering students plus class 12 students too....ur teaching is awesome ...thank you sir for ur amazing teaching skills and plz continue the series
The Lazy Engineer everything. Something about it that doesn’t get into my head. Thank you for replying as well. Idk how busy you are and taking the time to reply is very appreciated. I get confused with R2 to R3 mapping than I get even more confused with R3 to R2 mapping. Please if you can, make a video on it or several videos. The final I have is on Thursday.
The Berg abstractly, a linear map, A, is something that satisfies: A(c*u + d*v) = c*Au + d*Av. Where u and v are vectors and c and d are scalars. That’s it! That’s what defines linearity. Matrices actually come from this property. Ever wonder why we multiply vectors and matrices like we do? It all comes from this abstract idea of linearity (and basis vectors but we can hold off on that for now!) So if you have a map A from R2 to R3 all that means is that for a vector u in R2, when we apply A, then v=Au is in R3! A must be a 3X2 matrix to do this. I thjnk it helps to think of a matrix as a linear function that maps one vector to another. And we just happen to represent this linear function in the form of a matrix.
