YES! My fave subject! Linear algebra is all about to make our life easier by rotating and stretching everything until we find the best axis reference. Now it's time for the UV decomposition + a good explanation of why statisticians ( :) ) can't live without it (principal component analysis + minimax...even now google is using it with our data :( ). But first my all fave decomposition...which is like discovering the fire again 'cause every students starting to learn linear algebra can "guess" IT MUST BE LIKE THAT. I'm talking about the A=QS decomposition which, in the end, states that each and every single transformation matrix A (m x n) can be expressed as the unique product of an orthogonal matrix and a simmetric matrix! In other words, every linear transformation is just the composition of a rotation/reflection and some stretching! I found it amazing!
Do you know of any videos that explain this? I do pure maths at university but we don't really get taught the geometric interpretation of linear algebra, at least not yet.
Not only did you help me understand this topic, but by coincidence, this exact question with this exact quadratic equation was on my linear algebra final! Thank you sir!
There are just so many interesting paths to wander down beyond the typical curriculum. Thanks for sharing your enthusiasm and wisdom! Your channel is super fun, Dr Peyam!
The formula you were talking about, Dr Peyam, is simply the determinant of the quadratic form. Could you explain how that is, or what that is? If I remember my conic sections right, if the determinant0 then ellipse
That’s not quite true, the point is that you can tell the nature of your quadratic form by the sign of the eigenvalues. The condition you’re thinking of also involves the sign of the first term of the matrix, and is weaker than the condition I’m writing
Thank you Peyam, but I have a question. The original equation corresponds to iperbola that has y=x as major axis. And this should be correct, beacuse rotating the x-y axis counterwise by 45 degrees I get the canonical form of equation with new coordinates
At the end you checked what happens when x = 0. But the same happens for y = 0. The curve crosses y axis in 2 points as well. The graph you drew is rotated when compared to a graph Wolfram Alpha draws. Did you mix the axes on the final picture?
I found the video. While i see what you mean, i was wondering if there existed some matrix A (independant of x and y) which could apply to [(x),(y)] and contain some term like xy, or x^2, or if anything of the sort exists. does this matrix behave in a way that mimics such a thing?
you work with the first three terms, find the transformation, then change variables of the fourth and firth terms to the new variables and finally complete squares
@Patrick Salhany Well indeed it is also symmetrical about y = -x but it is not needed to graph the solutions of the equation. For the focal axis, I don't see what difference there is between the two since we also have a symmetry about x = 0 (or y = 0) wich transforms one into the other. The only thing that differs between the two could be a left handed / right handed version of the same object if you see what I mean. Chirality to precise.
@Patrick Salhany Oh I see, I guess on the picture he drew the focal axis was x = 0, but since the equation we have is not really a hyperbola eqution, either it has no focal axis, either we need another definition of the focal axis, am I right?
Could you use some similar technique to apply to cubic forms? But would you need some sort of 3 dimensional matrix then? Edit: And higher degree polynomials
To discover it’s a hyperbola, actually all that’s needed is the matrix (easy to construct) and the sign of the eigenvalues! Since the eigenvalues are positive and negative, it’s a hyperbola!
@@drpeyam Couldn't we just look at the upper left determinants of the original matrix? Well, I guess its the same thing, but that way you don't need the eigenvalues themselves?
I think he needed to plug y_2=0 into the new formula to see that the hyperbola branches away from y_1=+/-sqrt(7) in the new coordinate system +/-(sqrt(7)/sqrt(2),sqrt(7)/sqrt(2)) in the old coordinate system, so it should be opening towards the top-right and bottom-left.
@Patrick Salhany The only problem is that he drew the hyperbola in the x-coordinate system in the end. Correct would have been in the y-coordinate system.
@12:41 "even in higher dimensions or anything..." and my face breaks XD so then, with a 3x3 matrix we should be able to classify cubics rights? Finding the appropriate eigenvalues and eigenvectors in three space, Diagonalizing and Normalizing the new basis, This is what allowed you to list off the defining features of the curve, and like you said it would handler whatever translations to the axes were necessary for whatever case. Immediately I want to go to things like to folium of descartes, and factoring cubics in general. I'm sure cardano's formula for solving a cubic in x just falls out of the matrix multiplication somehow!?!? Does it relate in someway to Galois Theory or more simply why a quintic function wouldn't have a general formula to find the solutions?!?! Likely this would be an approach to proving things like " x^3 - xy + y^3 = 0 has no rational points other than the trivial ones" I really should have studied polynomial bases more in lin. alg. this is so fascinating! is this a topic in differential geometry? Always love your work Dr. Peyam Thank you for the upload!!!!
Oh, what I meant with the higher dimensional case is stuff like x^2 + y^2 + z^2 - 3xy + 2yz = 1 or something like that, which is easily generalized, but it doesn’t work with cubic terms
@@drpeyam ahh okay thank you! I should have realized, it must be a conic or at most two degrees in whatever variable (x^2 + y^2 + z^2 + etc. = 1) Thanks for the reply!
YES! My fave subject! Linear algebra is all about to make our life easier by rotating and stretching everything until we find the best axis reference. Now it's time for the UV decomposition + a good explanation of why statisticians ( :) ) can't live without it (principal component analysis + minimax...even now google is using it with our data :( ).
But first my all fave decomposition...which is like discovering the fire again 'cause every students starting to learn linear algebra can "guess" IT MUST BE LIKE THAT.
I'm talking about the A=QS decomposition which, in the end, states that each and every single transformation matrix A (m x n) can be expressed as the unique product of an orthogonal matrix and a simmetric matrix! In other words, every linear transformation is just the composition of a rotation/reflection and some stretching! I found it amazing!
Clap clap clap. Wonderful comment
Do you know of any videos that explain this? I do pure maths at university but we don't really get taught the geometric interpretation of linear algebra, at least not yet.
You helped me understand a very important topic in Matrices, and I can't thank you enough Dr. Peyam!
Not only did you help me understand this topic, but by coincidence, this exact question with this exact quadratic equation was on my linear algebra final! Thank you sir!
Wow what a coincidence!!
There are just so many interesting paths to wander down beyond the typical curriculum. Thanks for sharing your enthusiasm and wisdom! Your channel is super fun, Dr Peyam!
I think a similar technique is used to uncouple a system of coupled ODEs, which makes solving them much easier.
This video is perfectly related to my statistics. For us in statistics quadratic forms are very important.
Same
Thank you sir i actually used to solve this type of question in5-6 mins, after watching this video i am solving it in 10-20 secs.
Really liked your explanation! Please keep up the good work!
Love the way you're teaching. Thank you very much
I wish I knew this back when I was doing multivariable calculus in UNI :)
Excellent video buddy! keep going!
Judging from the eqn in terms of y1 and y2, the graph should be a hyperbola aligns with the new axes y1 and y2
Yeah
yes, I was looking for this comment :P
The formula you were talking about, Dr Peyam, is simply the determinant of the quadratic form. Could you explain how that is, or what that is? If I remember my conic sections right, if the determinant0 then ellipse
That’s not quite true, the point is that you can tell the nature of your quadratic form by the sign of the eigenvalues. The condition you’re thinking of also involves the sign of the first term of the matrix, and is weaker than the condition I’m writing
This guy sure loves science lol
Thank you Peyam, but I have a question. The original equation corresponds to iperbola that has y=x as major axis. And this should be correct, beacuse rotating the x-y axis counterwise by 45 degrees I get the canonical form of equation with new coordinates
I love your tune, it makes the lesson even more fun!
Thanx a lot sir
You really saved me in my semester exams....
Much thànkful tó you
Hi! I'm Jaime, a particle physicist. I love your videos! Thanks for posting. Just a comment: in this case, the hyperbola should cross the y_1 axis.
At the end you checked what happens when x = 0. But the same happens for y = 0. The curve crosses y axis in 2 points as well.
The graph you drew is rotated when compared to a graph Wolfram Alpha draws. Did you mix the axes on the final picture?
Maybe
I found the video. While i see what you mean, i was wondering if there existed some matrix A (independant of x and y) which could apply to [(x),(y)] and contain some term like xy, or x^2, or if anything of the sort exists. does this matrix behave in a way that mimics such a thing?
Sir after see your video with your confidential smile ......
Felling well for solve the problems of linear algebra
Thanks 😊😊
nice! a great way to simplify a quadratic.
Nice, but, generally speaking, how about the general 2D case Ax^2+Bxy+Cy^2+Dx+Ey+F=0 where the matrix X^T AAX is not enough to describe?
It reduces to my case up to a translation
you work with the first three terms, find the transformation, then change variables of the fourth and firth terms to the new variables and finally complete squares
And if you plug y = 0, you get x = +/- 1/sqrt(2), because the equation is symetric for x and y. So the graph is symetric about x = y eheh
@Patrick Salhany Well indeed it is also symmetrical about y = -x but it is not needed to graph the solutions of the equation. For the focal axis, I don't see what difference there is between the two since we also have a symmetry about x = 0 (or y = 0) wich transforms one into the other. The only thing that differs between the two could be a left handed / right handed version of the same object if you see what I mean. Chirality to precise.
@Patrick Salhany Oh I see, I guess on the picture he drew the focal axis was x = 0, but since the equation we have is not really a hyperbola eqution, either it has no focal axis, either we need another definition of the focal axis, am I right?
Love the way you teach... you are a great man
I learnt how to do this with curves and surfaces, even thought I eventually forgot it since I never got around to use it
Thanks for the video, but you drew the hyperbola as if is y1y2 = 1, rather than y1^2 - y2^2 = 1.
yes i thought so too...
As a software engineer, computing the value of 'y' does not save any work. But mathematically, it is cool.
So, if I had an application were I could compute one value of 'y' and then use it over and over again, that would be even more cool.
Thanks for helping me a lot !
You are a godsend.
This guy is awesome! New subscriber.
Thanks so much!!
Liked the video. Respect the excellent work!
amazing professor!
Nice video, but I did not get how you got the D-matrix suddenly.
Check my video on Diagonalize 2x2 matrix
This guy is so happy 😄
Could you use some similar technique to apply to cubic forms? But would you need some sort of 3 dimensional matrix then?
Edit: And higher degree polynomials
Tensors maybe? But I don’t know how to diagonalize them
Great content, as usual
Hello sir,
How did you draw the new coordinator system of y1,y2
The eigenvectors (the columns of matrix P) give the directions of the new axes.
Something doesn't add up: the equation is symmetric (you can swap x and y and the equation is still the same) but the graph isn't
Yes, his graph was simply wrong. The hyperbola is symmetric to the axis y1, not to the axis x2.
subscribed. Very clear explanation.
Can this method tell us something about the foci of the conic?
I really like this! But... it took 13 min without the diagonalization process just to discover it was a hyperbola. Maybe memorizing is not that bad...
To discover it’s a hyperbola, actually all that’s needed is the matrix (easy to construct) and the sign of the eigenvalues! Since the eigenvalues are positive and negative, it’s a hyperbola!
@@drpeyam That would do for a great video too!
Thank you Dr. Peyam!
@@drpeyam Couldn't we just look at the upper left determinants of the original matrix? Well, I guess its the same thing, but that way you don't need the eigenvalues themselves?
11:30 magic
Great video!!
Can u give an explanation about higer dimension case?
Shouldn't the graph be symmetric in y=x?
I think he needed to plug y_2=0 into the new formula to see that the hyperbola branches away from y_1=+/-sqrt(7) in the new coordinate system +/-(sqrt(7)/sqrt(2),sqrt(7)/sqrt(2)) in the old coordinate system, so it should be opening towards the top-right and bottom-left.
Here's what the graph really looks like:
www.desmos.com/calculator/awx5u4ea4u
@Patrick Salhany The only problem is that he drew the hyperbola in the x-coordinate system in the end. Correct would have been in the y-coordinate system.
@@philp4684 now it makes sense. the graph Peyam drew was too simple for something with xy term.
no wonder i thought something was wrong
I have drawn the hiperbola at Mathematica and it is different from your drawing.
Yes, his graph is wrong.
how do u get P???
@12:41 "even in higher dimensions or anything..." and my face breaks XD
so then, with a 3x3 matrix we should be able to classify cubics rights? Finding the appropriate eigenvalues and eigenvectors in three space, Diagonalizing and Normalizing the new basis,
This is what allowed you to list off the defining features of the curve, and like you said it would handler whatever translations to the axes were necessary for whatever case.
Immediately I want to go to things like to folium of descartes, and factoring cubics in general.
I'm sure cardano's formula for solving a cubic in x just falls out of the matrix multiplication somehow!?!?
Does it relate in someway to Galois Theory or more simply why a quintic function wouldn't have a general formula to find the solutions?!?!
Likely this would be an approach to proving things like " x^3 - xy + y^3 = 0 has no rational points other than the trivial ones"
I really should have studied polynomial bases more in lin. alg. this is so fascinating!
is this a topic in differential geometry?
Always love your work Dr. Peyam Thank you for the upload!!!!
Oh, what I meant with the higher dimensional case is stuff like x^2 + y^2 + z^2 - 3xy + 2yz = 1 or something like that, which is easily generalized, but it doesn’t work with cubic terms
@@drpeyam ahh okay thank you! I should have realized, it must be a conic or at most two degrees in whatever variable (x^2 + y^2 + z^2 + etc. = 1)
Thanks for the reply!
Hello. How could i find an orthonormal basis from quadratic form?
As he explained in the video: write down the matrix A for the quadratic form, calculate the eigenvectors of A and orthonormalize these eigenvectors.
how can i like a video more than one time
Really really cool. :-). Thanks !
you're the king! awesome
bro, what's your instagram ? i need you in my life
You made it seems so easy you are the best. Thank you very much
please do video about hermitian forms!
Why are P and D those matrices?
Look up how one diagonalizes a matrix.
You've drawn it as rectangular w.r.t. the new axes? Is that right?
No, that was wrong.
It cuts x1 and x2
thanks
i did it this way: 2x1^2+10x1x2+2x2^2=x1(2x1+5x2)+x2(5x1+2x2)=[x1 x2][2x1+5x2;5x1+2x2]=[x1 x2][2 5;5 2][x1; x2]
i like that the square root of a parabola is an hiperbola or half an hiperbola
very easy 2 follow TY
What was the horrible math text on Conic Sections?
All of them! But I think Stewart in particular haha
Merci beaucoup! bonne journée
De rien 😄
Wonderful as always :)
Merci beaucoup !
De rien! :)
Peyam got HELLA tan, cuz. He wasn't that tan before!
I got that tan from being stuck in LA traffic actually 😂😂😂
graph is wrong in the last
A is a matrix
b is a real number
How A can equal to b?
x^T A x is a real number
@@drpeyam it is 1×1 matrix.
Yep, which is the same as a number