literally Narcissa Squilliams
2 Likes
1 Like
the initial for the discord would be A, because A is better than B
1 Like
fuck that goes hard
1 Like
me personally S1E3 is my favorite
otherwise ur rankings are pretty good, assuming ur not saying that these episodes are bad just better than others
1 Like
No
1 Like
1 Like
I must apologise. This is outside my MENTAL realm
saying this anytime anyone asks me something i dont know
1 Like
I don’t think Arcane has any bad episodes: S1E2’s just meh because it’s both setting up a bunch of stuff and just repeating a lot of S1E1.
1 Like
similarity is determined not by characteristic and minimal polynomials, but by the Jordan Canonical Form. A matrix with char. poly (x - 3)^4 and min. poly. (x - 3)^2 has two possible Jordan Forms, one with 2 2x2 Jordan blocks and one with 1 2x2 and 2 1x1 Jordan blocks. matrices of these two forms are not similar to each other
1 Like
you’re all learning so much today
1 Like
Yep.
Learning a lot about how the mute function works.
4 Likes
so you see the reason why we dont really learn anything is that i have no clue whats the building blocks of that question
1 Like
this court finds you guilty of anti-cookie thread activity and orders you to be shot
2 Likes
inner product spaces might be more up your alley. it’s just a generalization of the dot product
1 Like
wait but how can the geometric multiplicity of the eigenvalue influence the structure of the Jordan Canonical Form in this case, and how does it relate to the number of Jordan blocks for a given eigenvalue???
3 Likes
jordan peterson?
3 Likes
that’s a good question! The number of Jordan blocks for a given eigenvalue is actually equivalent to the number of eigenvectors for that eigenvalue. while this is often the multiplicity, you do have to take into account that the matrix could be defective, and therefore have fewer eigenvectors for a given eigenvalue than its multiplicity.
3 Likes
This is to say, the JCF gives us information about how many total eigenvectors there are (so the first case in that problem there are 2 eigenvectors for the eigenvalue 3, and in the second case there are 3). either way, the matrix is defective, because 3 is the only eigenvalue, with multiplicity 4.
2 Likes