Woodbury matrix identity Puns

That’s okay, I always commute.

Where’s my Maths Nerds at?

During the infamous "Architect scene" in The Matrix Reloaded, the Architect tells Neo the truth about the history of the Matrix and his role as the One. Among other things: he reveals that the Oracle was originally created by the Machines to help him design an improved version of the Matrix after the earliest versions failed.

As he explains: the earliest versions of the Matrix failed because the Architect was created by the Machines as a purely logical being, giving him a severely limited understanding of human nature—which hampered his ability to create a virtual world that humans would actually want to live in. The Oracle was designed to be an empathetic being who could understand human nature in ways that the Architect couldn't.

As he specifically says:

"Thus, the answer was stumbled upon by another. An intuitive program, initially created to investigate certain aspects of the human psyche. If I am the father of the Matrix, she would undoubtedly be its mother."

Did you catch that? He reveals that the Oracle was originally created to investigate certain aspects of the human psyche to help the Machines understand humans better.

There's a word for someone who "investigates certain aspects of the human psyche" in real life. They're called a psychologist.

While the Architect doesn't spell it out explicitly, the implication is that the Oracle's "wise fortuneteller" gimmick was a later development, and that she was originally something more akin to a psychotherapist. And as we can clearly see in the reveal trailer for The Matrix Resurrections, Neil Patrick Harris' character is a psychotherapist.

Do the math. Maybe Neo's therapist is really the Oracle in a virtual disguise. Or maybe he's a new incarnation of the Oracle. Either way: considering the Oracle's history, it seems very likely that she and the therapist are connected in some way.

Hello,

I was wondering exactly what the title says. If a matrix A is guaranteed to be nonsingular, is there a matrix B s.t.

AB = I, B is NOT eq to A^(-1)

I'm mostly asking this because of my linear algebra homework lol. I like to verify if the inverse I found is correct, but multiplying out both sides AA^(-1) and A^(-1)A is tiring, so I was wondering if I have to truly do it both ways to confirm that it is the inverse ,especially since I've already found that the determinant is not 0.

Thank you!

In a PIM I got "-nan" in some areas instead of a number, and I can't for the life of me find what it means. Does anyone know? I thought it may have something to do with the fact that there are some seq where I got a negative branch length result (purposefully)

I have a great idea for this crossover, but I have no idea who should get what skin. Wanna help decide?

I was listening to Morph too much. And googling just to kill the time too much too.

Found N. Bourbaki mentioned in proofs for automorphisms – for instance:

https://preview.redd.it/rochylcc7sa61.jpg?width=800&format=pjpg&auto=webp&s=7abd93acc22580115a3a0596d3df10b7b7557acd

https://mathoverflow.net/questions/90964/lie-algebra-admitting-some-hyperbolic-automorphism-is-nilpotent

of which I understand nothing – I’m not good in Algebra (nor in English – quick apologies for people enjoying correct grammar – English is my second language, so please skip as there’s nothing of importance here :) it’s just this unresolved duality of dualism keeping me up

And the ones and zeros were bothering me. I tried looking for less advanced Algebra and found identity matrix and identity morphisms.

where I is the identity matrix (which is trivially involutory); R is an identity matrix with a pair of interchanged rows; [from Wikipedia of course]

Automorphism is a symmetry of the object, and a way of mapping the object to itself while preserving all of its structure.

The identity morphism (identity mapping) is called the trivial automorphism.

An identity relation or identity map or identity transformation, is a function that always returns the same value that was used as its argument.

the identity matrix is invertible—with its inverse being precisely itself [end of Wikipedia quotes]

Not sure if it makes any sense, but since many are now making wild guesses about the sixth album – my guess would be it might touch the subject of identity. I know it is more a wishful thinking from what I am personally working through just recently, but I just wanted to put it out there.

Also people in dema are described as devoid of any trace of an identity in Clancy’s letter. Maybe it could be the next step – not only fight and not only escape but work on what we allow to define ourself and why. I have no idea how to put in an understandable way – sorry.

I’m not sure if it was already discussed? If anyone can point me where I would appreciate. I know this is a long shot and based off practically nothing, but in case someone would like to have look into that - let me know what you think?

Every damn time I’m there, people already in the circle stop to let others in. I don’t understand how every driver I’ve seen there gets it wrong.

Hi all,

If I understand correctly, one difference between XMPP and Matrix is that Matrix is transported over HTTP, whereas XMPP is usually transported over TCP. It can also be transported over HTTP, with BOSH.

At first I wasn't planning on setting up BOSH. However, if I understand correctly, even when using TCP, Prosody requires a certificate to support encrypted connections.

I currently automate my Let's Encrypt certificates with a tool called cert-manager for Kubernetes. It's a lot easier than doing stuff manually, e.g. it renews certificates automatically for me. Thus I want to use it for Prosody. But due to limitations with Kubernetes, this isn't easy to do with a non-HTTP service. So I'm going to set up BOSH, let cert-manager create a TLS certificate for the HTTPS connections, but then also use that same certificate with Prosody.

Is using the same certificate for HTTPS connections (BOSH) and encrypted TLS connections a valid strategy? Or must they be different certificates?

I am also confused how the identity of my server will remain recognized, even when the TLS certificate expires and is automatically renewed. In contrast, Matrix has a matrix_key.pem, a private key that is strictly tied to the domain and doesn't change, even between different installations. Does XMPP have something analogous to this, or does it work differently?

Thanks for reading! Please feel free to point out anything I'm misunderstanding.

So i'm trying to solve a system of reactions as such:

A + B -> C

C -> A + B

C + D -> E

A + E -> A + F

to this describe this i have the following:

- a 1D matrix with starting values:

name_arr = np.array(['A','B','C','D','E', 'F']) # for printing/plotting
conc_arr = np.array([ 5,  3,  0,  0,  0,   0])  # starting concentrations

- two identity matrices: ( i need 2 since in the last reaction A is used and produced, so in a single matrix it's coefficient would be 0. but if there is no A to start with the reaction can't happen )

consume_arr = np.array([[1,1,0,0,0,0],
                        [0,0,1,0,0,0],
                        [0,0,1,1,0,0],
                        [1,0,0,0,1,0]
                        ])
produce_arr = np.array([[0,0,1,0,0,0],
                        [1,1,0,0,0,0],
                        [0,0,0,0,1,0],
                        [1,0,0,0,0,1]
                        ])

Multiplying the consume_arr with the conc_arr give me:

out_arr = conc_arr*consume_arr
>>> [[5 3 0 0 0 0]
     [0 0 0 0 0 0]
     [0 0 0 0 0 0]
     [5 0 0 0 0 0]]     

now i need to check the product per row, but only where the consume_arr is non zero. So out_prod_arr would be

>>> [15,
     0,
     0,
     0]

I figured out how to do the per row product:

np.product(out_arr,axis=1)

using the np.where function hasn't really helped either as it is returns a 2d matrix

i can do it easily by looping through the elements but i believe that loses any advantage of using numpy. so is there a way i could do this, or what are some good search terms to find the solution to this?

after this is i would multiply this product vector with the produce_arr, sum per column and get the new concentrations that way

The Hero of the War, John A. Dix

The man who saved the Emancipation Act, Levi Woodbury

", at the antipodes of the Republican model, which postulates the equality between human beings, independently of their characteristics of origin, sex, religion. It is the breeding ground for a fragmentation of societies that converges with the Islamic model"

Source is here: https://www.opendemocracy.net/en/can-europe-make-it/academic-freedom-in-the-context-of-frances-new-approach-to-separatism/

The source is crying about what the french government is saying and doing but I am quietly celebrating. If reform of Universities is happening in France then it can very easily spread to other countries too. No one can call Macron guilty of Trumpism.

so it would be like an inverted version of a positive angelic galactic origins reading exploring the reverse matrix monad/fallen star races and fallen angelic orders with which you are associated - rather than highest potential and most "high dimensional" past lives - these would look to the dark side and cosmic history of your godhead.

For more information DM me where I can provide more info about my other services and how to book.

The problem in the title is my paraphrasing of the second part of a past homework problem. In the first part, we had to prove that A having rank 'm' makes the linear function L_A :F^n -> F^m , given by L_A (*x*) = A*x*, onto. I had no problem doing that, and I figure I have to use that result in the second problem, but idk where or how. The T.A. tried to solve it in class, their argument was something like

&gt; "A being rank 'm' means that no row contains all zeroes, so each row has a leading nonzero value...(something, something)... so if a_ij is the leading nonzero scalar in the i-th row of A, define the b_ji-th element of B to be 1/a_ij , and let all other entries in the i-th column of B be zero, for all 1<=i<=m."

But the TA later said that their argument wasn't entirely correct, and in any case, I tried using the formula b_ji = 1/a_ij where a_ij is the leading nonzero element of the i-th row of A, and it didn't quite work correctly (though it was close), and I believe that it won't work whenever one column of A contains nonzero leading values for more than one row.

On my own, I haven't been able to find any theorems that really help, and I'm just lost on how to go about this proof. All I can really do is show that such a matrix B must exist, as there exist vectors *b*1, *b*2,...*b*m in F^n such that A*b*j=*e*_j since L_A is onto, where each *e*_j belongs to the standard ordered basis of F^m, and you can just define B as the matrix with column vectors *b*1, *b*2, etc.