Anonymous at Sun, 3 Nov 2024 10:55:17 UTC No. 16460776

>>16460756
Wtf does the symbol to the left of 0 mean?

Anonymous at Sun, 3 Nov 2024 11:53:50 UTC No. 16460817

>>16460776
NTA but that symbol usually means "is positive semi definite" in the language of semi definite programming.

So, if A is the matrix in the middle, then x'Ax>=0 for all feasible x.

Anonymous at Mon, 4 Nov 2024 00:37:20 UTC No. 16461598

>>16460756
I did my PhD in semi-definite programming
I can safely say I've forgotten every part of it so I can't help you

Anonymous at Mon, 4 Nov 2024 00:40:59 UTC No. 16461600

>>16461598
well, at least it got you a job

Anonymous at Mon, 4 Nov 2024 00:42:11 UTC No. 16461603

>>16460756
Trace = sum of eigenvalues
Det = product of eigenvalues
It's already in block form so you can treat the eigenvalues separately I think
One of the eigenvalues is 1-y_1, with eigenvector (0,0,1)
It's also symmetric which means its eigenvalues are either all real or one real and one complex conjugate pair (I think)
Just mess around with it
Just look up the wiki page for Block Matrix, it has all sorts of properties you can use

Anonymous at Mon, 4 Nov 2024 01:02:57 UTC No. 16461613

>>16460756
so, when talking about a positive definite matrix, the key property is that [math]\vec{v}^TM\vec{v} \geq 0[/math] for non zero vectors.
the motivation for this is the matrix acts like a generalized inner product, and in the case of applying a vector to itself the induced norm.

so try applying the matrix to a vector and see if you get any constraints.
in this case, [math]\vec{v}^TM\vec{v} = 2 v_x v_y y_1 + v_y^2 y_2 + v_z^2 [1 - y_1] \geq 0[/math]

Anonymous at Mon, 4 Nov 2024 01:57:35 UTC No. 16461653

Since it's in block diagonal form, the eigenvalues are the eigenvalues of each block separately. Notice that the block B is just a single entry.

Anonymous at Mon, 4 Nov 2024 02:00:20 UTC No. 16461654

>>16461600
Oh I don't have a job

Anonymous at Mon, 4 Nov 2024 02:11:24 UTC No. 16461663

>>16461654
yeah I dont believe you

Anonymous at Mon, 4 Nov 2024 02:14:54 UTC No. 16461667

>>16461598
Based. Linear matrix inequalities are gay anyways.

Anonymous at Mon, 4 Nov 2024 03:06:49 UTC No. 16461713

>>16461613
y1=0 is the only feasible solution. Otherwise that eigenvalue is negative.

Anonymous at Mon, 4 Nov 2024 05:15:49 UTC No. 16461812

>>16461713
Correct, the determinant of the first block in
>>16461653 is equal to -y_1^2. Thus only y_1 = 0 is allowed. The matrix then collapses into diagonal form with eigenvalues 0, y_2, and 1. Overall a fairly boring example

Anonymous at Mon, 4 Nov 2024 07:24:12 UTC No. 16461875

>>16460756
Work out the Determinant and the Trace. You'll see that the solution is obvious

Anonymous at Mon, 4 Nov 2024 07:25:07 UTC No. 16461876

>>16460756
use the gauss-jordan method

Anonymous at Mon, 4 Nov 2024 15:33:01 UTC No. 16462222

>>16461598
>>16461654
Based