Thank you. That's what we aim for

Thanks for your feedback - much appreciated.

Thank you - glad to hear that!

Glad you liked it!

Looking for the proof of this property as well...

Hi Tom, Very generally speaking (the devil is in the details) the eventual connection will be that if you take a stable system (positive eigenvalues in the second variation of energy) and perform "small oscillations" dynamic analysis, you will end up with a system of differential equations with minus the matrix and therefore minus the eigenvalues.

Kudos for embodying the mantra - never stop learning, no matter the age!

Tlad you enjjoyed it!

Hi Rishabh, Great question! There's a flaw in the argument. If ABC = RST, it doesn't mean that A=R, B=S, and C=T. (Consider I*I*I=A^(-1)*A*I.) Hope this helps! Keep up the great questions! Pavel

Lemma Really Thanks for your reply and i clearly understand what's the point you are making here and i am agree with that but Prof. Gilbert Strang has done this type of comparison among two matrices in one of his video and the name of the lecture is SVD where he compares A^tA = V(E^tE)V^t with eigen value decomposition. Can you help me how does he is doing that or there is something which i am missing ?

Lemma This is the link of that video kzfaq.info/get/bejne/o6iTf7V42qeqhZ8.html

@@rishabhgarg9217 You are wrong. The Prof. Gilbert Strang has not done the comparison rather has obtained the expression from A^tA. regards

Hi Joshua, An extreme example would be the identity matrix. Are its eigenvectors orthogonal? Depends on how you choose them, but you can always choose them to be orthogonal.

@@MathTheBeautiful Okay, wait a sec...I think I understand this! Actually, whether I understand it or not depends on your answer to this question: If a symmetric matrix has two eigenvectors that get scaled by the same eigenvalue, then does that matrix have an entire circle of eigenvectors that gets scaled by the same eigenvalue? (Please say yes....) In other words, are the only possible 2 by 2 matrices with 2 eigenvectors with the same eigenvalues diagonal matrices? What about 3 by 3 matrices? How would we get two eigenvectors with the same eigenvalues in that case? Thanks!

First question: Yes. Second question: Diagonal matrix with 2, 2, and 3 on the diagonal

​@@MathTheBeautiful And I suddenly woke up today and realized why this makes sense. Lol, it's actually much simpler than I thought. Let's say we find two eigenvectors that get scaled by the same eigenvalue. That must also mean (simply by the way vector addition works - say we have two vectors, a and b, and some constant k, then if a + b = c then ka + kb = kc) that every possible linear combination of those two vectors get scaled by the same eigenvalue, but doesn't change directions. In other words, every single possible linear combination of those two vectors is an eigenvector of the matrix. And, since the two original eigenvectors point in different directions, we'll be able to linearly combine them to create an entire circle of same length vectors that get scaled by the same eigenvalue as those two vectors do, and from that circle we'll be able to PICK two eigenvectors that are perpendicular!

@@joshuaronisjr Correct! That's why it's called an eigenspace.

chagrin

@@MathTheBeautiful Thanks. I never get tired watching this series of lectures.

Bücherregal Domi lf so what is its significance in geometry. it is that wormholes which is door to all 10 dimensions .

you use cool words that are not "chagrin" to me.

What the ⟂ are you talking about?

@@Hythloday71its auto generated word by Google captions not by him.

Hi Rob, I know how you feel! Just wanted to let you know that there's absolutely a way to ask for help. I've created an online system to enable people to ask questions. Check out https:/lem.ma/LA1 where you'll find that you can ask a question in every video and every problem. I'm sure you'll love every aspect of the platform and the content. Pavel