@@theengineerguy8726 well said. I always start a new topic from 3b1bs videos, get excited, then confused and then return to Dr Treffor to find the intuition.
A geometric explanation for what you talk about at 11:40 is that if all column vectors have a zero at the N-th coordinate, then they are 'locked' on a single plane and cannot 'rise' from it to the 3d space. More generally, for an N-dimentional space, if all vectors have a 0 at a fixed coordinate, then they only live in a space one dimention smaller, R^(n-1). And any vector not lying on their plane is unreachable, because it lives in a higher dimention - that is, no solution for `x` will take you from A's column vectors to `d`.
Have no words ❤️ , actually Im trying to understand his every bit word . For this video I took much time and I saw it 6 times . But worth doing it was really amazing
12:08 I wonder if we could make this part a little more rigorous "if Ax = b has a solution for every b∈R^m then Rx = d has a solution for every d∈R^m". The constant vectors b and d will not necessarily be the same due to of row operations. I think this follows from the fact that row operations on b only scale or add multiples of other b rows, so the generality of the 'every' isn't lost.
Truly excellent and far, far clearer than the much more popular Khan and 3B1B series.
Thank you so much!
Nope 3B1B is great.
both are great@@saravanabalajik
the 3b1b series is really well made, but its supposed to be an advanced course, its really just a cool introduction to the field.
@@theengineerguy8726 well said. I always start a new topic from 3b1bs videos, get excited, then confused and then return to Dr Treffor to find the intuition.
Your explanations are very edible for unprepared minds. Thank you from Moscow. Great job!
A geometric explanation for what you talk about at 11:40 is that if all column vectors have a zero at the N-th coordinate, then they are 'locked' on a single plane and cannot 'rise' from it to the 3d space.
More generally, for an N-dimentional space, if all vectors have a 0 at a fixed coordinate, then they only live in a space one dimention smaller, R^(n-1).
And any vector not lying on their plane is unreachable, because it lives in a higher dimention - that is, no solution for `x` will take you from A's column vectors to `d`.
I am starting from this video in the playlist, and very excited. I love it when math shows how things are equivalent.
You are my actual professor for this class and I can't thank you enough. My "professor" does't even lecture.
Glad I'm able to help, but the prof you paid for should be doing it!
Well simple, ...and very easy to understand... Thank you sir for making Algebra simple.
Thank you so much!
Have no words ❤️ , actually Im trying to understand his every bit word . For this video I took much time and I saw it 6 times . But worth doing it was really amazing
Finally! From here it begins to be practical for exams!
12:08 I wonder if we could make this part a little more rigorous
"if Ax = b has a solution for every b∈R^m then Rx = d has a solution for every d∈R^m".
The constant vectors b and d will not necessarily be the same due to of row operations.
I think this follows from the fact that row operations on b only scale or add multiples of other b rows, so the generality of the 'every' isn't lost.
Sir you are a HERO
just wondering if the proof will still hold if there is a free variable?
12:00 What about infinite solution case , in that case there will be no leading 1 in one of the column?
The proof for #4 is Proof by Contrapositive. That is, one way to prove that #3 => #4 is to prove that not #4 => not #3.
well explained!
THIS VIDEO IS WHAT MAKES YOU DIFFERENT FROM THE TYPICAL MATH TEACHER
It will be really great if you could post some quizzes for each videos to test the understanding professor!!
What a bummer! The videos are great, I'm binging them currently. Would be even more helpful with quizzes
🔥🔥🔥
Why was d video cut short ?
im so confused what do the m and n stand for? I thought they meant m (rows) and n (columns)
They are, but in the linear combination form. This of a1, a2... as matrices of m rows, there are n of them>>> we have a m*n matrix
thank you so much