In college I never had a robust course on Linear Algebra (as an engineering student we played with matrices a little, not much though). This is so disappointing given the power of the subject, it's reach, and mathematical beauty - thank goodness for the internet!
A very good lecture, but there is a slight mistake. At 4:05 you should multiply with the conjugate transpose of y rather than just transpose, as the eigenvectors of a symmetric matrix can be complex even though the eigenvalues are always real.
I'm not an expert and I'm just getting into advanced linear algebra myself, as I am in my sixties, but I thought that in order for the system to be stable, The characteristic polynomial of the matrix, had to result in negative real eigenvalues or complex eigenvalues with a negative real part, as opposed to positive eigenvalues. Where am I wrong?
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.
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!
@@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!
Can i also prove that eigenvectors of Symmetric Matrices Are Orthogonal in the following way ? As we know any symmetric matrix can be decomposed into A= LDL^t and from eigen value decomposition A=XEX^-1 where E is eigen value matrix, so by comparing these 2 equations X=L, D=E and L^t = X^-1 which implies that X^-1 = X^t which is the property of orthogonal matrices. Thus X is Orthogonal matrix . Is this a right way or not ? Thanks in advance.
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 ?
Wow. As someone who is math challenged but is very curious to learn, this lecture is like my WORST NIGHTMARE. Droning on and on about theory which I'm not quite following and then a launch into demonstration which I also, don't quite follow. I feel like I SHOULD be understanding what's covered here, but in reality I'm not and there's no way to ask for help. *MATH* *ANXIETY* *x10000%!* No offence to the lecturer, I'm sure It's my issue as a student, but, YIKES!
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
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
He sure is proud of his symmetric matrices
Man you are a phenomenal professor, you really deeply understand the material. Thanks for the great lecture
Beautiful. Concise yet comprehensive
Thank you. That's what we aim for
This is great. I'm working on research now, and I have used your videos two times now to quickly understand key concepts. Thanks!
In college I never had a robust course on Linear Algebra (as an engineering student we played with matrices a little, not much though). This is so disappointing given the power of the subject, it's reach, and mathematical beauty - thank goodness for the internet!
This is EXACTLY what I was looking for like for 2 hours
Crazy good! Thanks for the content, I find it very educational and well represented.
Thanks for your feedback - much appreciated.
I wish you were my lecturer. My lecturer is great but so are you!
you are so good in linear algebra
Thank you very much! I was looking for how to prove the eigenvectors are orthogonal so hard, and finally I got the answer here!
Thank you - glad to hear that!
Thanks a lot sir for this beautiful proof.
Glad you liked it!
Very Good!
thanks to dr for your lecture
A very good lecture, but there is a slight mistake. At 4:05 you should multiply with the conjugate transpose of y rather than just transpose, as the eigenvectors of a symmetric matrix can be complex even though the eigenvalues are always real.
brilliantly simple.
I'm not an expert and I'm just getting into advanced linear algebra myself, as I am in my sixties, but I thought that in order for the system to be stable, The characteristic polynomial of the matrix, had to result in negative real eigenvalues or complex eigenvalues with a negative real part, as opposed to positive eigenvalues. Where am I wrong?
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!
Thanks 🙏
Tlad you enjjoyed it!
WARNING: Gratuitous use of orthogonal symbol ! !!
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.
How can we prove the dimension of the eigenspace equals the multiplicity of the corresponding eigenvalue?
Looking for the proof of this property as well...
Thank you!!
Thaaaaaank youuuuu!!!!
What did he say at 2:32 ?
chagrin
@@MathTheBeautiful Thanks. I never get tired watching this series of lectures.
What do you mean by "the eigenvectors can be CHOSEN to be orthogonal if the eigenvalues are the same"?
Thanks!
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.
The bit in the video we accepted without proof. Is there a proof on this channel for it now? I can't find it
awesome
please sir send this question as soonas If A is non singular operator than spectrum ofATA inverse equal to spectrum ofT
Can i also prove that eigenvectors of Symmetric Matrices Are Orthogonal in the following way ?
As we know any symmetric matrix can be decomposed into A= LDL^t and from eigen value decomposition A=XEX^-1 where E is eigen value matrix, so by comparing these 2 equations X=L, D=E and L^t = X^-1 which implies that X^-1 = X^t which is the property of orthogonal matrices. Thus X is Orthogonal matrix . Is this a right way or not ? Thanks in advance.
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 ruclips.net/video/mBcLRGuAFUk/видео.html
@@rishabhgarg9217 You are wrong. The Prof. Gilbert Strang has not done the comparison rather has obtained the expression from A^tA.
regards
Wow
Ohhhh baby
I've learned fuck all.
Man I love Eigenvectors so much
Bücherregal Domi
lf so what is its significance in geometry.
it is that wormholes which is door to all 10 dimensions .
Wow. As someone who is math challenged but is very curious to learn, this lecture is like my WORST NIGHTMARE. Droning on and on about theory which I'm not quite following and then a launch into demonstration which I also, don't quite follow. I feel like I SHOULD be understanding what's covered here, but in reality I'm not and there's no way to ask for help. *MATH* *ANXIETY* *x10000%!* No offence to the lecturer, I'm sure It's my issue as a student, but, YIKES!
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