Quick Cauchy Schwarz Proof

In this video I provide a super quick proof of the Cauchy-Schwarz inequality using orthogonal projections. Enjoy!

[@ParthVerma-kc1wr]

I love how happy he gets when he teaches!! I hope all teachers were like this haha

[@Newtonissac6]

Man, you are the biggest life saver ever. Just now, I was doing all these inequalities in Hilbert spaces and I have been stuck on this one particular problem for a while now. I am gonna go grab some popcorn. Let the video begin...

[@mridultyagi2995]

I was solving sammer Bansal book for jee advance . And there was a problem mixed with calculus and this inequality and your video helped me . thanks

[@user-xb1vc9no6x]

I really like the way you talk. I can feel your passion about math even though I am only watching the video!

[@drpeyam]

Thanks so much!!!

[@aminahanjum3592]

studying for lin alg final was very down bc my prep didnt feel good enough but seeing your video made me happy thanks for teaching with enthusiasm

[@TVSuchty]

This is a cool easy and short proof. Very elegant.

[@narayandhakal3526]

I'm also left handed ✋️ 😅 Thank you for making this video!

Man I love you, the way you make these problemas that easy is just incredible!

[@bosepukur]

your videos are wonderful , need more of such excellent work

[@elchaflan100]

manage to keep a smile while explaining, i dont know if i should take him serious

[@JoaoGabriel-eb9nd]

No, no one smiles doing math, he’s a reptilian no cap

[@joelvirolainen590]

Rewinding rules! I love these videos!

[@AndDiracisHisProphet]

Man, I remember when I was a little Erstsemester I thought "What is this stupid thing good for!? " Was I naive at that time :/

[@justwest]

AndDiracisHisProphet Do you say "Erstsemester" in english? Is that actually a word, like Kindergarten?

[@AndDiracisHisProphet]

Nicht, dass ich wüsste

[@lin1450]

@@AndDiracisHisProphet xDDD

[@renesperb]

A very simple proof is as follows : consider the quadratic function Sum(u+t*v)^2 (1 to n)> = 0. Then the discriminant condition is just the Cauchy-Schwarz inequality . The same way it follows in a Hilbert-space if you replace the sum by the general scalar product.

[@drpeyam]

I did that one already

[@FranqiePR]

Thanks Dr. Peyam!

[@lucifer.333]

please continue making such videos🤙👍

[@ap6872]

Nice sessions Sir I am from India 🇮🇳...

[@yuvalpaz3752]

I don't remember this proof, but this inequality has so many ways to prove it you can't expect from one to remember them all. Well, in Euclidean space this come straight out of the definition of dot product(as |cos(x)|

[@cameronspalding9792]

If I was proving this: I would consider which is a quadratic function in t which is non negative hence has a non positive determinant

[@drpeyam]

Check this out: Cauchy Schwarz Proof kzbin.info/www/bejne/iYGminaMlpp3o68

[@sachi5427]

the best proof of cauchy schwarz i've seen!

[@drpeyam]

Thank you!!

[@cancermelon2155]

Cool! I also like the proof made by Serge Lang (maybe it's the standard proof idk) If v = 0 easy. If v ≠ 0, let x = (v • v) , y = - (u • v). Then, since: (u x + v y) • (u x + v y) ≥ 0 dividing by: (v • v) we get: (u • u)(v • v) - (u • v)^2 ≥ 0 QED

[@drpeyam]

Yep, that’s also a nice proof!

[@stydras3380]

that indeed was quick

[@mkhan28.]

Thank you so much. ❤️

[@taubone9257]

Neat! Link to the "swag" proof?

[@drpeyam]

Proof of the Cauchy-Schwarz inequality kzbin.info/www/bejne/iYGminaMlpp3o68

[@Noporeh2]

Interesting, i'm used to prove the triangle inequality using CS but you're doing it the other way around. So i'm curious, how do you prove the triangle inequality ?

[@FranciscoTornay]

You don't need the general triangle inequality for this The proof relies only on the special case of a right triangle, which is obvious from Pythagoras' theorem

[@jabunapg1387]

From the definition of ||x||

[@Zoonofski]

You legend!

[@demudu]

Very cool

[@flooreijkelboom1693]

Love this proof

[@EL-eo8dh]

Very cool!

[@YorangeJuice]

Nice!

[@danieledwin8927]

At 2:02 Shouldn’t it be the projection of u onto v rather than u hat? Sorry if I’m watching some dodgy version of the correct thing or if I made a mistake

[@drpeyam]

u hat is better, especially if it’s clear what you’re projecting on, the proj notation is lengthy and confusing

[@Rolljack]

THANK S a lot

[@EmissaryOfSmeagol]

Would saying that |u*v|

[@drpeyam]

That only works for vectors in R^2, my proof is more general

[@PuzzleQodec]

Makes me wonder if the concept of an angle could be generalized to higher dimensions, or to vector spaces in general. Would it make sense to 'define' an angle theta between u and v as the inverse cosine of |u∙v| / ||u|| ||v||?

[@ryliur]

At 0:50, when v is represented as a zero vector, does that not also apply for u?

[@drpeyam]

u and v are independent of each other, so if v = 0, u can still be anything we want

[@ryliur]

@@drpeyam Sorry I may have worded my question a little awkwardly. If we set u = 0 instead of v, would we also be "done" with the proof? I was just slightly caught off guard by the emphasis of setting *only* v to be a zero vector.

[@drpeyam]

I see! Yeah it doesn’t matter, you could set u = 0, but then in your second case you would divide by u and not by v. It’s basically symmetric

[@ryliur]

@@drpeyam Thanks for the explanation!

[@ricbisa6092]

13 people didn't understand the proof.

[@drpeyam]

😂😂😂

[@myonlynick]

2:00 u-hat how come to be the formula as shown? i Know that u-hat= ( u / ||u|| )

[@drpeyam]

That’s a different u^, my u^ is the projection of u on v, check out my video on orthogonal projections

[@ElizaberthUndEugen]

||û|| = cos(u, v) ||u||, ok? then we have ||u|| ||v|| cos(u, v) --------------------------- v ||v|| ||v|| ||v|| cancles once so we have ||u|| cos(u, v) --------------------------- v ||v|| Now v/||v|| is a unit vector in direction of v and ||u|| cos(u, v) is the length of the projection of u onto v (in other words ||û||, as stated above). Thus v ----- ||u|| cos(u, v) ||v|| is û.

[@AndDiracisHisProphet]

Hey, Peyam. I have to apologize. I made STEVE call you, to (ab)use you as a voice-over app. I am sincerely sorry that he stole your precious time. :/ . . . . Naaah I'm just kidding, that was hilarious :D Next time teach him how to properly pronounce "Weierstraß" :D

[@drpeyam]

Hahaha, it’s ok, it was really funny and a great idea!

[@MiroslavMakaveli]

yes lets do some limits

[@MiroslavMakaveli]

they do computer science and have done math

[@123bubbleup]

You are adorable

[@drpeyam]

🥰🥰🥰

[@aaryan505]

R U Indian?

[@zacharieetienne5784]

Slightly flawed proof. What if u is in the span of v? Trivial case though

[@drpeyam]

If u is in the span of v, then u = c v and so v = (1/c) u, so v is in the span of u (and c is nonzero since u is nonzero)

[@zacharieetienne5784]

+Dr. Peyam's Show Well yes, but that doesn't prove much. You've just gotta do the same thing you did with v = 0, except substitute u = cv. Both sides should be equal and thus less than or equal to.

[@drpeyam]

Sorry, misread your question! It’s actually a not a flaw in the proof, but a feature of the proof! If u is in the span of v, then not only is the formula valid, but moreover u^ = u, so we get equality! And more than that, this is the *only* way we get equality in the C-S inequality! Neat, huh?