Proof of the Cauchy-Schwarz inequality | Vectors and spaces | Linear Algebra | Khan Academy

Proof of the Cauchy-Schwarz Inequality

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[fanrco]

I highly recommend trying to prove this theorem in R2, R3, and R4 just to get a better intuition on how the variables cancel out. My proffessor gave us a take home test where we had to prove cauchy schwartz in R4 and it was honestly a very beautiful proof. The day after in class he was able to use the structure for the proof in R4 to generalize to Rn

[boipelojoe]

b/2a to minimize the function --- it's a quadratic function; it achieve its minimum at b/2a... "Let me pick any number, b/2a" -- we know that's not any value :) You can't prove it without b/2a
You can also differentiate p(t)
From (y.y)t^2 - 2(x.y)t +x.x
p'(t) = 2t||y|| - 2(x.y) *** x.x is a scalar
p'(t) achieve its min when = 0
=> 2t||y||^2 - 2(x.y)= 0
=> t||y||^2 = 2(x.y)
=> t = (x.y)/||y||^2
**Substitute t; the proof is complete.

[revooshnoj]

I think Sal failed to explain a key insight here. First lets break down P(t)=|| ty-x ||^2 Notice that ty is a parameterization of a line so ||ty-x|| ^2 is essentially the squared distance between a point on that line and a vector x. Now what we want to do is find the minimum distance between the vector and the line. Since the function P(t)= at^2-bt+c obtains a minimum at t=b/2a (notice the negative sign in the function) we substitute that in to get the minimum distance and get the inequality using thr fact that P(t)>=0 (all lengths are positive) .
Note:If ty and x are on the same line their minimum distance is zero which explains the equality and why one must be a multiple of the other

[possumsam]

at 8:17, the intuition for using b/2a is because t at b/2a is the minimum point of the curve. By completing the square, we
get a ( t - b/2a )^2 - (b^2)/4a + c. Since a is positive, t = b/2a is at the minimum point of the curve. So the reason for choosing b/2a is to evaluate the inequality at the point closest to 0. Can anyone confirm or deny?

[Mariomation]

you are my motivation. I think your videos are more of complements to classes rather than actual lessons. it so diverse and inclusive that if you were to study a topic complete your still guaranteed to learn something new.

[Augustine_]

Very interesting, but I've got one problem. From where did you get this p(t) artificial function?

[jpfry]

Awesome proof!

I think an argument with the discriminant would be a little more natural, rather than plugging in b/2a which takes a bit of foresight and might seem arbitrary.

For those asking, the Cauchy-Schwarz inequality is extremely useful in Linear Algebra, Analysis, and probability. A good amount of proofs in mathematics use this inequality (even in physics too, where CS is used to arrive at the Heisenberg uncertainty principle).

[thetruereality]

Is there some specific reason why you defined P(t)=||ty-x||? Can you define any other vector differently?
Also later in the proof you evaluated p(t) at b/2a you have not mentioned what is the a reason for that?
It was a nice explanation I would be grateful if you explained above questions.
Thank you

[sdfsgdsfgsdfg]

8:06 could use some motivation for t=b/2a

[rinkaghosh]

Why choosing the function p(t) ? And then t=b/2a ? It might be better if some purposes were told for choosing the function and then that particular value of t

[JaySomething]

I think that it's more convenient to just use the identity that the dot product of any two vectors is their magnitudes multiplied together, times cosine of the smallest angle between them, to prove the inequality, or the equality when they are parallel vectors since theta (smallest angle) is 0, so cos0 is 1. For any other non-parallel vector pair, since costheta has the domain between neg 1 and pos 1, then we divide both sides by the magitudes to get 1 >/= costheta, which is true.

[energysage]

Why not just say A dot B = |A||B| cos(theta), and the largest value cosine can contribute is 1, giving A dot B = |A||B|. Cos(theta)=0 occurs at 0 and Pi radians (or 0 and 180 degrees), meaning A and B must be parallel or antiparallel. Either way, this means they are scalar multiples of one another.

[rikard]

how do I know what constant to put in to P(t) to prove the theorem? It seemed like you knew the answer and you used the "answer"(the constant t which was b/2a)) to "proof" the theorem.

[Furkan-yvew]

Since I don't really know the definitions i can't prove them by my self. And I can't find a starting point for this. Is it necessary to know some linear algebra to understand topology

[borissimovic]

Where does the P(t) function come from, what is the intuition for using it?

[amyluo]

Thank you soo much for giving us free lessons!
Education is the most valuable thing ever.

[jakubnage]

Hello Salman Khan. I love your videos and have used them for calculus. You are truly an asset and valuable resource to any student.

The main purpose of this comment (if you see it) is to ask what you recommend I use with my wacom bamboo tablet whilst taking notes in class. In terms of functionality, my only prefernce is that I could have typewritten text together with my tablet writing which would be diagrams.

Anyone who feels like answering this, comment here or message me

[jingyiwang]

I am really grateful for this video! You have provided a really clear explanation about this inequality. Thanks! My mathematics exam is approaching. I am so nervous about it.

[pjgcommunity]

The equation at(squared)-bt+c in the video is very similar to the equation to find distance after t seconds when acceleration is an arbitrary constant(1/2at(squared)+bt+c)

[Ashkepu]

Where is the equation p(t) = ||ty-x||^2 from? How did you choose it?