Linear Independence

[@blackpenredpen]

Abby!!!!!!!!!

[@drpeyam]

Hahaha

[@isharauditha4257]

That is you behind 🌝

[@drewpat9535]

Dr. Peyam, like your colleague, you're extremely clear and your smile is a benefit for the followers. But, please, consider shrinking your board, cause it's a very tough task to reay at the corners!!!

[@drpeyam]

I’ll try!

[@bkramkoirala]

@ Dr. Peyam I was looking at the solution of the following question. How to determine linear terms from the nonlinear dataset? I will make it more clear by taking one example. Let us take the parametric curve r(t) = [t×t;t], t = [0,1]. Using this equation, I generate 1000 points. Now my goal is to determine the value of t for each point on the curve without using the equation of the parametric curve. To solve this problem, I estimated the geodesic distance from the starting point (t=0) to all points on the curve. The initial idea was to relate these distances to the barycentric coordinates. In this case, it is t. But the result is not satisfactory. Is there a way to determine the linear term (t for this case) by just using the dataset itself?

[@MrRyanroberson1]

honestly 9:00 is asking "are there a and b when c = -1"

[@mortamor2352]

so I just did linear independance in school so I came up with this: for the sake of writing it more easily: [1|5|-3] = A ; [-2|-9|6] = B; [3|h|-9] = C then I used those vectors to get this equation: A = bB + cC where b and c are Real numbers then I got a system of equations by using the x1, x2 and x3 of those vectors so then it looked like this: I: 1 = -2*b + 3*c II: 5 = -9*b +3*h III: -3 = 6*b - 9*c then I solved I for b and got b = (1+3c)/-2 which I plugged into II and got c = 1/(2*h+27) the I plugged that back into II to get b in terms of h and got b = -(h+12)/(2h+27) with b and c solved I came to the conclusion that the solution is h e R \ {-27/2} Please tell me if that is correct!

[@DutchMathematician]

The way I look at it... Let's take a general matrix A. Elementary row operations do not change the dimension of the row space of a matrix (which is relatively easy to prove and most textbooks on Linear Algebra will prove/show this). (the statement above also holds if we replace every instance of 'row' by 'column') The fact that elementary row/column operations do not change the dimension of the column/row space (in that order!) is a bit less trivial. But if we take for granted that dim(row space of M) = dim(column space of M) for every matrix M (also often proved/stated in textbooks), then this is 'trivial'. Now take the particular example in the video: / 1 -2 3 \ A = | 5 -9 h | \ -3 6 -9 / A simple row-reduction of A gets us the (row) echelon form of A: / 1 -2 3 \ A' = | 0 1 h-15 | \ 0 0 0 / (note: the echelon form of a matrix is NOT unique: you might get another matrix!) From the form of this matrix A' we can immediately conclude that dim(row space of A') = 2 (forget about 'pivots', just look at the rows!). Since elementary row operations do not change the dimension of the row space of a matrix, we can therefore conclude that dim(row space of A) = 2. But since dim(row space of A) = dim(column space of A), we know that the three columns of A only span a 2 dimensional space. It is a 'well' known fact that the set of columns of A that correspond to pivot columns in A', are linearly independent. Hence, we know that the first two columns of A are independent. Since we already know that dim(column space of A) = 2, the third column of A must be a linear combination of the first two (independent) columns of A. However, how do we find such a combination? Well, reduce A' further to what is called 'row reduced echelon form': / 1 0 2h-27 \ A'' = | 0 1 h-15 | \ 0 0 0 / (note: the row reduced echelon form of a matrix IS unique: you should get the exact same matrix!) It turns out that the following holds: for every non-pivot column in the row reduced echelon form of A, the numbers in its column exactly describe how the corresponding column of A can be written as a linear combination of the columns of A corresponding to the pivot columns of A'' (note: this only holds for the row reduced echelon form, not for the reduced echelon form!) In this case: 3rd column of A = (2h-27)*(1st column of A) + ((h-15)*(2nd column of A) In the video they took the example for h=1, but the formula above gives the general expression.

[@nootums]

Hey Dr.Peyam!! Absolutely love your videos!! Can you please make a video on how to solve homogeneous system of matrices. When does it have no solution and when does it have infinitely many solutions? Thanks!!

[@drpeyam]

It never has no solutions since 0 is a solution. It has infinitely many solutions if there isn’t a pivot in every column

[@nootums]

@@drpeyam By no solution I meant trivial solution 😁😁

[@drpeyam]

Trivial solution if there’s a pivot in every column, nontrivial if it isn’t

[@manceaugael7960]

Can you explain the Bell polynomials Bn,k please I don't understand it

[@drpeyam]

I don’t know what those are

[@manceaugael7960]

Ah I see, it seems like really few people know about them

[@youssefbenhachem993]

Guys can u help me in this integral : Integral from 0 to 1 of 1/(1-x) + 1/ln(x)

[@MartinPuskin]

Apparently it equals the Euler-Masceroni constant, cool!

[@edwardhuff4727]

Yes it converges, but it's not obvious why. The antiderivative is Integrate[1/(1 - x) + 1/Log[x],x] = -Log[1 - x] + LogIntegral[x] See also Numberphile on this constant. kzbin.info/www/bejne/apyUm5idipmKmJY at 6:30.

[@youssefbenhachem993]

Oh yeah guys , i know that it is equal to the euler constant but can u prove it ?

[@ElifArslan-l9g]

thank you:)