Diagonalize 3x3 matrix
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- เผยแพร่เมื่อ 6 ก.ย. 2024
- Diagonalizing a 3x3 matrix. Finding eigenvalues and eigenvectors. Featuring the rational roots theorem and long division
Check out my Eigenvalues playlist: • Diagonalize 2x2 matrix
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His enthusiasm makes this feel like an exciting adventure.
adorable teacher, effectively learning, Thank you!!
You’re welcome! 😄
In soviet russia 14:12, equation long-divides you.
He is such a happyyyy man!!! 😂😂 I wish I could be this much happy too while doing linear maths! 🤣
This is a great explanation, thank you!
"Heeeeey, but it's so much easier to factor out (lambda+1) here and (lambda+1) here!"
(3:34)
Isn't it supposed to be 4 in the first column instead of -4?
Good tutorial on diagonalization. Thank you
Thank you very much. You are very effective and cool teacher.
love your attitude!!
love your enthusiasm about lin algebra
If you normalize each eigenvector to unity, the matrix P will be orthonormal and its inverse will equal its transpose. So no work beyond normalizing the eigenvectors is required to get P inverse.
I love this guy's energy
The RRT is fairly easy to prove: Let f(x) be a polynomial of nth degree with integer coefficients. Assume p/q is a rational root of that polynomial. Then f(p/q) = 0. If you multiply that equation by q^n, all terms on the left hand side will be integers. Of these, the leading term a_n p^n has the distinction of being the only term that is not a multiple q. So we can subtract it, then bracket out q on the left hand side, and we get that -a_n p^n = q(some integer). Since p and q are coprime, the only way that equation can hold is if a_n divides q. That's why the denominator must be a divisor of the leading coefficient.
Returning to our equation f(p/q) * q^n = 0, we see that what is left of the constant term a_0 q^n is the only term that is not a multiple of p. Analogous to the above, it follows that p must be a divisor of a_0, so the numerator must be divisor of the constant part.
For a more generalized case of cubic: You should use make the function into a depressed cubic, then solve it by comparing the depressed cubic with the identity (m+n)^3, then jump into the complex world that gives you some sort of cube root of a complex function, and you one of the solutions, the other solutions could be found with long division.
But if you do this then this video will be like 10 hours
Wow thank you, you made me understand how to do this, you explained it very well, but i didn't understand at all how you did the determinant, anyway i tried it using other method and i got the same solutions!! Thank you very much
Greetings from Spain!!
my best reagrades and respect from egypt
You have a strong positive vibe💪
If you set lambda to 10, you can use number theory to perform a prime factorization. In special cases, this will lead to a factorization of lambda when converted back into a variable. A prime example of this is x^2+2x+1. This converts to 121 which factors as 11^2. 11 = x+1. For certain factors of P a*10 + b might not convert easily back to ax+b because while 5*6 = 3*10, (x-5)(x-4) \= (x-7)(x). I’d be interested in a video on the conditions where the factor ax+b = a*10+b.
came to learn about digonilization of matrix learned amazing fact about polynomial
But I thought that the rational roots theorem says, that if you're taking an exam, all roots of a cubic polynomial are integers between -3 and 3.
And also there is one brutal formula that directly calculates the characteristic polynomial p(x).
p(x)=x^3-trace(A)*x^2+(det(A1)+det(A2)+det(A3))*x-det(A).
Ai is the minor of A which is acquired by bomberman-ing the i-th row and column.
Tant qu'on y est, on aurait même pu calculer l'inverse de A en utilisant le théorème d'Hamilton-Cayley ^^
J'ai hâte de voir la suite ! : ) On a un peu de trigonalisation de prévu ?
that sign mistake solution was nice.
he is super excited. great video.
Nice exercise! I calculated P-1 to be [0 -1 1,1 2 -1, -1 -1 1] in your columnorder where the commas seperate the rows. Thanks
it is so useful for me!!
i wish you were my professor
❤️
Congratulations for the work. The matrix of the video cover is wrong. I tried to solve without looking at the solution and came up with a complex solution.
Thanks for the video
Welcome :)
When you see 3x3 matrix, you are hoping for Symmetric positive definite. Makes things so much easier...lol.
Definitely LOL
Hello i really enjoyed the video but i have a doubt if an eigenvalue has 2 basis what will be the P matrix then ?
So this is basically using Horner's method to find the Lambdas
Thank you so much for your videos ! Very clear and good energy 😊😀
Thank you. Great video.
Great Work
Thank u sir...♥️
u are welcomed
Thank you very much lecture if possible you may explain for Me more and many exercises and I need to attend this class
Amazing
Thanks sir
It is a great explanation but you made it complexly no need for all that ,teacher
you look really happy lol
god bless you
Why not factor by grouping at 6:30?
great vid👌
Thank you very much sir
The row reduction method seemed to me longer than simply multiplying out the vector (x, mx + c ) and then quickly solving 3 simultaneous equations. You still end up at the same place don’t you?
Never heard of this other method :)
how did lamda times - 3 equal a -4lamda ?????
Show the time so we can find the issue
Thank u sir ❤️
What is the condition for a matrix to be diagonalizable?
I don't know about necessary conditions for the general case, but there are some important sufficient conditions: if the matrix is symmetric or Hermitian, it's always diagonalizable
Basically enough eigenvectors :)
Is the diagonal matrix unique for the other matrix?
Determinant non-zero
my teacher wants it for a 3x6 matrix where all the values are in the couple thousand except for 1 zero and im dying
Omg I’m so sorry! Also 3x6 is not possible, do you mean 6x6?
@@drpeyam Here was the question: Let A be the 3 × 6 matrix given below:
Find invertible matrices P and Q such that P AQ is a diagonal matrix with
only 1s and 0s along the diagonal.
I said there was no solution since invertible matrices have to be square, which wouldn't produce a square output.
You could use Ruffini's Rule for that polynomial division?
What’s that?
Sounds like Fubini, haha
@@drpeyam Its basically synthetic division
@@drpeyam en.m.wikipedia.org/wiki/Ruffini%27s_rule
@@drpeyam it's very useful. We learn it at high school as an easy way to probe some number is root of a polynomial
Is this really so much fun for you my man?
Yes it is!
Why divide by lambda - 1 during long division?
It’s because 1 is a root, so lambda-1 is a factor and hence we can divide by it
@@drpeyam thanks for clearing that up. I was a little confused too.
Great video!!! FYI The thumbnail matrix does not match the video matrix. There's a 3 in the thumbnail where there is a 4 in the video.
Haha, clickbait 😂😂😂 But thanks, I’ll fix it
@@drpeyamHaha! Thanks!
How about synthetic division showing only coefficients?
I think that works, I’ve never learned synthetic division, actually
If a matrix is diagonizable then it has eigenvalues?
Yeah
Hello! I like a lot your videos and I would like know if you can make a video of triangulization with T-conductors, minimal polinomial, etc
Thanks a lot for too much math
I start laughing you when you are excited about finishing. I thought the finding the eigenvalue followed the formula A-λI and not reverse, or does it matter?
Doesn’t matter, since we’re setting it to 0
Can you speak some what slow it understands better resy of all it was awesome thankyou so much
You can always play the video at half speed :)
damn, really?
❤
10:54
olaf teaches linear algebra
RedPenGreenPenBluePen :P
Who encountered the annoying guy of amazon black Friday ads
He looks like Alex Aiono😍😅
Wow what a compliment!
professor, i say it again. You are cute and i cant focus on question due to that 😤😤
Jordan normal form pls
There’s a video on that already
Pro strats: use cubic formula @.@.
Kya paglo ki tarah bk rha hai
First uwu
Make it simple.soo boring
I have a 2x2 version
@@drpeyam don’t listen to him. You’re video are so exciting I eat popcorn when I watch them😊
Amazing