Linear Algebra 2i: Polynomials Are Vectors, Too!

[hussainmd]

Thank God I found this series, I'm taking linear algebra at college, but I was confused a little by the different aspects of the course. However, your teaching approach is much more attractive, understandable, and organized. A big thank you from Saudi Arabia.

[doodelay]

I genuinely love how you add context videos to your lectures and don't just hit us with definitions lol These are probably my favorite vids. I've made a chart below of how Linear Algebra connects with other fields and objects such as polynomials. I am not able to prove every connection so some of it is conjecture on my end. Nonetheless:

Linear equations -> Matrices -> Vectors -> Slopes -> Tangents
Tangents -> Curves -> Continuous Functions -> Polynomials
Polynomials -> Calculus -> Derivatives -> Integrals -> Series -> Limits
Continuous Functions -> Reals -> Complex plane.

So here we see that linear algebra leads to polynomials through the fact that vectors, treated as tangents, can describe continuous curves

[UnforsakenXII]

Haha, wow. I didn't realize this is what they meant by polynomials was a vector space. Thanks.

[antonellomascarello]

0:48 : Polynomials are vectors in the sense of LA
2:42 : #1 : Polynomials form a vector space
3:22 : Important Note : Polynomials of degree UP TO N (to restore the vector space property)
5:49 : #2 : Polynomials being "stuck" in a subspace (geometrically inspired concept)

[edgara]

This is a fantastic video for those that appreciate mathematics. Thank you very much for quality explanation.

[DEEPAKSINGHcreative]

There are examples in co-ordinate geometery about famliy of circles or lines. I don't know if they fall into same cateogory, but defination seems to be quite similar. Example of general polynomial passing through would be: (x-1)*(ax^n + ). So even if you add or subtract using linear combination logic still, the term (x-1) is taken common and hence whatever the new polynomial defination represents, it would still be passing through x = 1.

[erinb]

This is lovely, I'm so happy to have found this. <3 Thanks!

[chiveferdinard]

given that p1(x) = 5x^3+3x^2-2x+6, p2(x) = x^3+4x^2-3x+1. find p1*p2

[tangolasher]

+MathTheBeautiful

By the laws of linear algebra, you cannot multiply two vectors together, whether they are geometric or polynomial; only add them together in different proportions.

This makes sense for geometric vectors even outside linear algebra, but not for polynomials, which you most definitely can multiply - just add the exponents, so x^2 and x^3 would give you x^5, right?

So why the artificial restriction in linear algebra? Why can't I just multiply one basis "vector" by another to get a polynomial of the desired degree?

[BoZhaoengineering]

hi, Professor. I was wondering if your tensor textbook is offered in Kindle or PDF edition.

[MathTheBeautiful]

Arnau, sometimes "formal" means two opposite things. When I say "not formal", I mean not based on postulates and technical proofs. The concept of a vector space is very important and can be understood without postulates.

The formal approach has its value and is the right approach for many people, although I often feel that it teaches a formalization of the subject and not the subject itself. And the word "rigor" is often used to mean "technical". I see my approach as informal and rigorous.

[jarrodmccarthy]

"..particularly important in applied mathematics and physics where finding a simple answer in terms of a simple expression to a complicated problem is still considered the highest art; it wont be for much longer..."

what do you mean by "it wont be for much longer"?

[JimPIckins]

Are there exercises or quizzes to test comprehension of the material?

[NutritonFacts]

Hi prof, how to evaluate the output of the polynomial for particular input x0 if we represent polynomial as vectors ?

[rogerhom]

Something I'm not clear about in the last example - you demonstrate that polynomials with 1 as a root cannot be composed to form polynomials that do not have 1 as a root. However, that's not quite enough to show that a polynomials with 1 as a root are a subspace in the linear algebra sense. Is it possible to find a set of polynomials that form the basis of this subspace? If so, how many polynomials would we need for the basis?

[jonathanr]

idk why the quality on the lemma website is so bad. can't read much that you have written on the board.

[slapadashable]

I do see the similar properties which vectors share with polynomials but I still fail to visualize how polynomials can represent n-dimensional vectors? Maybe my understanding is narrowed to the n-tuple representations of n-dimensional vectors- please help!

[xoppa]

8:58 why did you erase and say that it is now a polynomial

[unperrier]

Being stuck on a plane doesn't sound that bad when you consider the list of in flight movies

[brianbertness]

I didn't come to this video for the LONG intro which consists of telling me what I already know about polynomials. The introduction to your video's subject should be at most 10% of the time spent. In this case it is much, much too long.