It says there's only 2 views right now. I'm so glad I get to see Dr.Peyam's videos so early. Your channel is simply one of the best and most informative math channels on all of RUclips.
This is my favorite theorem in all of math. It really was a surprise to me. When my professor presented it, I said "is that weird? That seems weird." He said "yes, it's surprising." On a side note, that professor (Arturo Magidin) got his PhD at Berkeley, small world. 😂
@@drpeyam Yeah, I went double check to make sure I didn't lie to you. He got his PhD in 1998, his advisor was George Bergman. He was the best professor I ever had, and was an absolute mathematical machine. He was also the coordinator and department head of our graduate program, so I spoke to him pretty frequently.
supposed your proposition at 1:20 is true, and A = PDP^-1. AA - 5A + 6I = 0 is true if and only if P(DD - 5D + 6I)P^-1 = 0. Next, notice P^-1 0 P = 0 for all P, and therefore the equation holds with or without reference to P. Next we take one important property of diagonals: Each number along the diagonal is totally independent, and therefore this can be treated as a system of equations where D = [a1,a2...], and we get that ai^2 - 5ai + 6 = 0. Therefore all elements along the diagonal are equal to some solution to what we would otherwise call the characteristic polynomial of A.
Sometime I tease young mathematicians with this question: How many roots for a 2 degrees equation ? And the good answer is an infinity number, thanks to Caley-Halmilton ....
@@drpeyam Makes me wish I could continue on with Masters (they stopped doing masters loans when I applied :( ) but Linear Algebra is still my main thing :D keep it up Dr Payem and Happy Holidays!!
There is a way involving trace of powers of the matrix to get characteristic polynomial without calculating eigenvalues tr(A^m) = Σ λ^{m} and we have sums of powers which is symmetric function but we want elementary symmetric functions because they are present in Vieta formulas Newton-Girard identities gives us connection between elementary symmetric functions and sums of powers and recursive version of Newton-Girard identities is enough to us When we have recursive version of Newton-Girard identities all we need to do is use Vieta formulas Cost of this approach is O(n^4) time complexity and O(n^2) space complexity but we dont need calculate eigenvalues nor use parameters in calculactions so it is quite easy to program it (There may be some numerical problems when we want to write code for it) Cayley Hamilton theorem states that matrix satisfies its characteristic equation ---------------------------------------------------------------------------------------------------------------------------------------------- His picture looks like an egg
Are T invariant subspaces the only time in math we ever use invariant to mean closed under something? Because here T invariant means the subspace is closed under the operation of turning x into T(x)?
Much simpler proof: just plug the matrix A in lambda whilst in form det(a - lambda*I) to get det(0) which is obviously 0 But seriously, pls explain why this doesn't work or give me a time stamp where he explains why this doesn't work
Because that doesnt make sense. The expression det(xI_n -A) is a polynomial function p : F -> F. in other words you have a zero matrix on one side of equals sign and a determinant(scalar) on the other. its like saying 1+2 = blue. Further if you input that expression into the definition of determinant, you get a matrix with matrix entries in the diagonal. I hope this helps.
Nice video. But I gotta say watching your video is the same level of difuculty as reading the text.You just give hint and we need to fill in the blank. Good for student studying pure math though.
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I cannot do this in my head.
Believe in the math!
"It might be a long video: my apologies". Absolutely no apology needed: keep them (the long videos) coming!
This is the most underrated theorem in all of math.
Love this. My linear algebra class ended right as we just began to discuss eigenvalues and diagonalization.
It says there's only 2 views right now. I'm so glad I get to see Dr.Peyam's videos so early. Your channel is simply one of the best and most informative math channels on all of RUclips.
Thanks so much ❤️
You sir, are what a truly teacher does: MOTIVATES
Thank you very much. I was in expectation to see the proof of this theorem explained in a didactic way on RUclips! Finally got it.
"Sorry if I sound tired but I'm okay" You're too precious🥺🥺, please take a rest if need to~
This is my favorite theorem in all of math. It really was a surprise to me. When my professor presented it, I said "is that weird? That seems weird." He said "yes, it's surprising."
On a side note, that professor (Arturo Magidin) got his PhD at Berkeley, small world. 😂
Haha, yeah, everyone is surprised by it 😂 And coool! Do you know when he got his PhD?
@@drpeyam Yeah, I went double check to make sure I didn't lie to you. He got his PhD in 1998, his advisor was George Bergman. He was the best professor I ever had, and was an absolute mathematical machine. He was also the coordinator and department head of our graduate program, so I spoke to him pretty frequently.
Oh cool! Yeah, way before my time :P But I know George Bergman, he’s cool!
Thank you so much for explaining so enthusiastic! You make my day :)
You’re welcome!!! 😄
Thank you a lot for your work 😇
supposed your proposition at 1:20 is true, and A = PDP^-1.
AA - 5A + 6I = 0 is true if and only if P(DD - 5D + 6I)P^-1 = 0. Next, notice P^-1 0 P = 0 for all P, and therefore the equation holds with or without reference to P. Next we take one important property of diagonals: Each number along the diagonal is totally independent, and therefore this can be treated as a system of equations where D = [a1,a2...], and we get that ai^2 - 5ai + 6 = 0. Therefore all elements along the diagonal are equal to some solution to what we would otherwise call the characteristic polynomial of A.
No, sorry, your proof is oversimplified. As usual, watch the video...
Sometime I tease young mathematicians with this question:
How many roots for a 2 degrees equation ?
And the good answer is an infinity number, thanks to Caley-Halmilton ....
haha nice one ^^
Nice proof. Were you thinking about doing any videos on tensor products? It seems like something you would do videos on
This makes me so happy because proving the CH Theorem was the end result of my Undergrad Dissertation using exterior products :D
Coooool!!!
@@drpeyam Makes me wish I could continue on with Masters (they stopped doing masters loans when I applied :( ) but Linear Algebra is still my main thing :D keep it up Dr Payem and Happy Holidays!!
Great work D peyam
ten times better explained than the proof my professor did...
Well done!
Thanks! 😊
Thank you so much. By the way I like your vibe :)
I really like his energy
I love you man
Nice video! :)
What book should I look at to get into the details of the proof of the theorem and the lemas you have shown?
Friedberg Insel Spence
@@drpeyam Thank you! :))
There is a way involving trace of powers of the matrix to get characteristic polynomial without calculating eigenvalues
tr(A^m) = Σ λ^{m}
and we have sums of powers which is symmetric function but we want elementary symmetric functions because they are present in Vieta formulas
Newton-Girard identities gives us connection between elementary symmetric functions and sums of powers
and recursive version of Newton-Girard identities is enough to us
When we have recursive version of Newton-Girard identities all we need to do is use Vieta formulas
Cost of this approach is O(n^4) time complexity and O(n^2) space complexity
but we dont need calculate eigenvalues nor use parameters in calculactions so it is quite easy to program it
(There may be some numerical problems when we want to write code for it)
Cayley Hamilton theorem states that matrix satisfies its characteristic equation
----------------------------------------------------------------------------------------------------------------------------------------------
His picture looks like an egg
Are T invariant subspaces the only time in math we ever use invariant to mean closed under something? Because here T invariant means the subspace is closed under the operation of turning x into T(x)?
So are there other things in math where Invariant is used where we would expect closed to be?
One can prove this result by working in the albebraic closure of the field
Is this theorem taught in high school in USA?
It’s taught in university
If I remember correctly, this theorem can be proven directly using adjugate matrices
Yes. And it is far quicker.
but in my mind not as intuitive
Much simpler proof: just plug the matrix A in lambda whilst in form det(a - lambda*I) to get det(0) which is obviously 0
But seriously, pls explain why this doesn't work or give me a time stamp where he explains why this doesn't work
Because that doesnt make sense. The expression det(xI_n -A) is a polynomial function p : F -> F. in other words you have a zero matrix on one side of equals sign and a determinant(scalar) on the other. its like saying 1+2 = blue. Further if you input that expression into the definition of determinant, you get a matrix with matrix entries in the diagonal. I hope this helps.
just to clarify, that proof may be simpler but it is not correct.
The proof is trivial if A is diagonalizable, but the surprising result is that it holds even if it's not, so you can't put A in the form lambda*I
Same in our IIT JEE notes so simple
Called characteristic equation
It blew my mind :/
Nice video. But I gotta say watching your video is the same level of difuculty as reading the text.You just give hint and we need to fill in the blank. Good for student studying pure math though.
The problem is that if I filled in the details, the video would be 2 hours long :)
Can I talk to you sometime :)