The Vandermonde Matrix and Polynomial Interpolation

[@robmarks6800]

Lovely! Would like to see a continuation into the fourier transform matrix!

[@nymiantoft5907]

That was so elegant

[@af9466]

Thanks for the video, this was a very interesting demonstration of an application of Vandermonde matrix I'd never heard of before, and all the steps were clear.

[@area51xi]

You went from stating a rule regarding rows but then the example used columns which is very confusing.

[@xizar0rg]

The framework is essentially the same as what's in linear algebra textbooks (slightly more handwaving on the arithmetic than what I remember) from back in the day but presented more succinctly. (got tired of having to code it anew in fortran every time I needed it... damned kids and their libraries for everything.)

[@martinepstein9826]

Great video. The Vandermonde determinant is really cool but I don't think it was necessary to show invertibility. If we know that Lagrange interpolation or whatever works for every choice of vector y that means the Vandermonde matrix is surjective, hence invertible.

[@General12th]

This is a nice little video!

[@yaniv242]

at 7:15 when you substract a cols, i cant seem to understand why the second row after it only has x0 in the power of 1, shouldnt the action of sub x0^2 on the above col affect the whole col? sorry that part got me confused a bit

[@jurrich]

Note the blue text: we are subtracting "x0 * the previous column", for all rows. First we look at the effect that has on the first row, namely that it turns every element after the first into a zero. Then we look at what it does for the second row, so we start with [1, x1, x1². x1³, ...] and then apply same operation: the 1 stays the same; the next column gets "x0 * previous column = x0 * 1 = x0" subtracted, yielding x1 - x0; the next column gets x0 * x1 subtracted, yielding x1² - x0x1; the next column gets x0 * x1² subtracted, yielding x1³ - (x0 * x1²), and so on. The next row [1, x2, x2², x2³, ...] gets the same treatment, turning into [1, x2 - x0, x2² - x0x2, x³ - x0x2², ...], and so on all the way down the matrix.

[@plushiie_]

omg you saved me!

[@DrWillWood]

glad it was useful! :-)

[@augustusnero5592]

This was the best explanation I've seen on the Vandermonde determinant.

[@DrWillWood]

Thank you very much!

[@TESLA_13_DR]

@@DrWillWood noo, this is the best explanation video I have ever seen on yt, bravo!

[@neolord50pro77]

@@DrWillWoodit’s the best explanation in the universe

[@drioko]

what a long and complicated explanation. i don’t understand this and this video is making me panic

[@pnachtwey]

Yes, this is long winded. I just write the 4 equations and solve for the unknowns. This is simple if the points are equally space. Sometime one must use unequal intervals like t01 for the time between x0 and x1. I get equations in terms of time intervals. I make the substitutions for the time intervals so they aren't calculated over and over again.

[@1495978707]

Cool vid! 3:30 I would like to note that, while for a proof, inversion of the matrix is probably the best way to go about this, practically speaking, inversion of the matrix is usually the worst way to go about solving a matrix equation. The idea is that if a matrix is invertible, A is unique, not that it exists at all. But if A isn’t unique, that means the interpolating polynomial isn’t unique. Anyhow, this matrix depends only on choice of node location, not value at the node, whether the equation has a unique solution depends only on whether the matrix is invertible, not on the y values. And that’s why it’s useful to do the proof this way.

[@slavinojunepri7648]

Your comment is absolutely correct. The matrix equation can never results in my multiple solutions for A because of the polynomial interpolation uniqueness. Therefore, the polynomial doesn't exist if the Vandermonde determinant is zero. Using the Gauss-Jordan method would be best to solve for A than having to invert the Vandermonde matrix. Matrix inversion requires the use of cofactors, a very computationally expensive and wasteful process. I cannot recall another way for doing so that would have an acceptable time complexity, say polynomial for instance.

[@mb59621]

For representation of the general formula of vandermonde determinant , a double pi notation could be easier to understand , like a nested for loop. The left subscript is the lower limit, and right subscript is the upper limit here .. j=1π(n-1) {i=0π(j-1)} (xj - xi) is the correct representation then .

[@dmitrypolozkov1335]

Thank you! Great video, keep new uploads on!! Greetings from Russia❤️🧸

[@richardfrederick1885]

The audio was terrible!!

[@pianoforte17xx48]

Finally the determinant vandermonde after searching literally everywhere with no hope of understanding it. You are a legend

[@djridoo]

Very cool ^^

[@kafkayash2265]

Would really like to see an explanation regarding sylvester matrix too. Your videos are really great!

[@joaoheleno3700]

You're beautiful

[@redroth57]

Awesome video

[@henrik3141]

Nice video. Just a small error at 0:53 to write that P_n = .... You also missed the chance of proving the "Key fact" by the vandermont determinant.

[@DAIQRY]

Amazing video!

[@BSK_666]

Very clear, thank you so much doctor..

[@uamdbro]

Nice, I always saw the existence proof given by explicitly constructing a polynomial (using Lagrange polynomials) and then using Taylor polynomials or something in actual practical applications. Though now that I am thinking about it, if all we want is an existence proof, doesn't that follow immediately from uniqueness? Seeing as the Vandermonde matrix (once you have fixed the x-values) represents a linear map on a finite dimensional vector space, where injectivity is equivalent to surjectivity?

[@nikita_x44]

Proof of uniqueness assumes existence. if it exists, then exists at most one.

[@surajchess3114]

But it doesn't say if a node is repeated then polynomial doesn't exist or multiple polynomials exist... so how to find the polynomial if a node is repeated? I can repeat the method for non repeating nodes and then multiply by (x-xr)^m when xr is repeated node and m is no of repetitions. Is this correct?

[@derendohoda3891]

If a node is repeated then you have two output values for a single input which is not a function and so not a polynomial. So what you are doing is not interpolation. You are looking for polynomial regression, probably.

[@AJ-et3vf]

Awesome video! Thank you!

[@the_nuwarrior]

its quite similar to the DFT matrix

[@bentationfunkiloglio]

Love it

[@rauldurand]

Very cool! What software do you use for the animation?

[@DrWillWood]

thanks! I use Apple Keynote for the animations

[@callmedeno]

Wow that was peak clarity, you are gifted at this

[@DrWillWood]

Thank you! Appreciate that! glad it was useful

[@kafuu1]

Nice work!

[@DrWillWood]

Thanks!

[@thespiciestmeatball]

That was awesome! Great job, Dr. Wood