Quadratic Forms

In this video, I use linear algebra to find the conic section 2x^2 + 10xy + 2y^2 = 1. The advantage of this approach is that it requires no memorization and that it works for higher dimensions as well.

[srinivaspavan]

You helped me understand a very important topic in Matrices, and I can't thank you enough Dr. Peyam!

[nirmalkarthik]

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!

[Zonnymaka]

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!

[notjohn]

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!

[suryodaybasak]

Really liked your explanation! Please keep up the good work!

[MrCigarro]

This video is perfectly related to my statistics. For us in statistics quadratic forms are very important.

[ugurburan]

Love the way you're teaching. Thank you very much

[nikshayyadav]

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.

[MushookieMan]

I think a similar technique is used to uncouple a system of coupled ODEs, which makes solving them much easier.

[_DD_]

I wish I knew this back when I was doing multivariable calculus in UNI :)
Excellent video buddy! keep going!

[kohimson]

I love your tune, it makes the lesson even more fun!

[nandapeela]

Thanx a lot sir
You really saved me in my semester exams....
Much thànkful tó you

[shwetathakra]

Sir after see your video with your confidential smile
Felling well for solve the problems of linear algebra
Thanks 😊😊

[ethancheung]

Judging from the eqn in terms of y1 and y2, the graph should be a hyperbola aligns with the new axes y1 and y2

[jhoefkensj]

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.

[hamayoonshah]

Love the way you teach... you are a great man

[madhuragrawal]

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 determinant<0 then hyperbola, =0 then parabola and >0 then ellipse

[prester]

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

[dhunt]

nice! a great way to simplify a quadratic.

[kipu]

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?