Diagonalize 3x3 matrix

Diagonalizing a 3x3 matrix. Finding eigenvalues and eigenvectors. Featuring the rational roots theorem and long division

[Peter_]

His enthusiasm makes this feel like an exciting adventure.

[hihiirokane_gbf]

adorable teacher, effectively learning, Thank you!!

[MrRyanroberson]

In soviet russia 14:12, equation long-divides you.

[tinyasira]

He is such a happyyyy man!!! 😂😂 I wish I could be this much happy too while doing linear maths! 🤣

[ralfschmelcher]

This is a great explanation, thank you!

[samg]

Thank you very much. You are very effective and cool teacher.

[mahreenathar]

Isn't it supposed to be 4 in the first column instead of -4?

[deborahodion]

Good tutorial on diagonalization. Thank you

[Furiac.]

love your enthusiasm about lin algebra

[elmer]

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.

[dominicellis]

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.

[vitalsbat]

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

[shokan]

but it's so much easier to factor out (lambda+1) here and (lambda+1) here!"
(3:34)

[nullplan]

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.

[MazenFiki]

my best reagrades and respect from egypt

[Rundas]

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).

Ai is the minor of A which is acquired by bomberman-ing the i-th row and column.

[AttiyaAnwarAhmed]

Hello i really enjoyed the video but i have a doubt if an eigenvalue has 2 basis what will be the P matrix then ?

[jeremyb]

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 ?

[buttersalad]

that sign mistake solution was nice. <3

[aaron_mg]

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!!