A Group of Uninvertible Matrices

In this video we'll show that a certain set of uninvertible matrices forms a group under matrix multiplication.

Music:
Deku Palace
Composer: Koji Kondo
Platform: Nintendo 64

Outro:
"Lateralus" as performed by Sakis Strigas
Originally by Tool

My current equipment for making videos (affiliate links):

[danielortega]

Hi! Is this some new channel? Just saw it in recommended videos and seemed nice, and it was.
Let's denote the matrix of all "a" by [a].
We have [a]•[b]=[nab].
Consider the isomorphism
g:A_n-->R\{0}, g[a]=na for all [a] in A_n. Clearly this is an isomorphism.
Now we have
So g is a group isomorphism.

[JM-usfr]

Just look at the group {2, 4, 6, 8} under multiplication modulo 10. Bizarrely, it's a group with identity 6.

[foliation]

The isomorphism to the real numbers maps each matrix A in the group to 2x in R (where x is the entry repeated 4 times in matrix A)

[Supremebubble]

Also a real lifehack for groups: You don‘t need commutativity for reasoning that you only need to check onesided neutrality and existence of onesided invertability as long as you show the same side each. It holds that: Given (G, •) with

1) • is closed and associative,
2) There is a leftneutral element e, such that ex = x
3) For every x in G there is a y such that yx = e

then (G, •) is a group. You can even exchange left for right, as long as you do the same side for both 2) and 3). This makes showing that something is a group slightly neater ;)

[jackcarr]

I'm new here and I'm really enjoying your channel. The information is presented neatly, and it's much more satisfying watching you write on my screen than on some stupid blackboard XD. It's presented neatly and everything is really well-explained. Plus, you manage to make it fun without any memes, which is really interesting - I've never seen this done before. I wish you luck with your channel!

[JamalAhmadMalik]

THAT WAS CRAZY!
I have been following you for just over a month, every one of your videos makes me think the same!
You are a REAL CONTENT creator!

[alvinlepik]

The group you demonstrated is a subsemigroup (not even a submonoid) of the 2x2 matrices under the usual matrix multiplication. In the sense of universal algebra semigroups, monoids and groups are completely different algebras. In order to speak of sub-structures, it must have all the same operations of all arities as the superstructure. Therefore, (A, x) is a subsemigroup, which turns out to be a group.

[AndrewDotsonvideos]

Sorry I didn't catch your mistake earlier. I think you meant to say "invertiblen't" instead of uninvertable. Rookie move.

[ownagesniper]

Love the group theory stuff. Keep going! You're helping me rediscover my passion for math!

[SloomFusion]

Wow you explained so well .. I didn’t find that obvious at start !! Thanks for the video 💪

[berserker]

Zeldaaa :3

Edit: anddd Lateralus to top it all off! You are a man of culture, respect.

[coconutflour]

Awesome video! I love groups and abstract algebra!

[Intrebute]

It's been a long while since I've done any homework. This was fun!



Here's the answer to your excercise, which is to exhibit an isomorphism between (R\0, *) and (A, *)
Notation:
[a] : 2x2 matrix with all entries a. i.e.
[ a a ] = [a]
[ a a ]
in this notation, the operation of A would then be written
[a][b] = [2ab]
Define the function f from R\0 to A as f(x) = [x/2]
This is a homomorphism because
f(x)*f(y) = [x/2][y/2]
= [2xy/4]
= [xy/2]
= f(xy)
and
f(1) = [1/2]


Define the function g from A to R\0 as g([a]) = 2a
This is a homomorphism because
g([x][y]) = g([2xy])
= 2*2xy
= 4xy
= 2x * 2y
= g([x])*g([y])
and
g([1/2]) = 1/2 * 2 = 1


They are inverses because
f(g([a])) = f(2*a)=[(2*a)/2] = [a]
and
g(f(x)) = g([x/2]) = 2* (x/2) = x

[yaboylemon]

Bro you are out here growing in subs good shit

[spsathiyapriya]

Thank you Professor. I m in India. I am looking for the method to solve non-invertible matrices ...Your video is of great use to me. 👍

[lukostello]

Could you please teach the mathematics behind Jeremy England's Formulas? I don't even recognize some of the symbols.

[benjaminbrady]

I have a question about matrices. I remember hearing that you couldn't take the inverse of a non-square matrix but if you have an nxm matrix then isn't there a mxn matrix that will leave you with the identity in mxm form?
For example, I'm fairly sure the inverse of:
[2 0 5]
[0 3/5 9]
Could be:
[1/2 0]
[0 5/3]
[0 0]
Which would result in the 2x2 identity
[1 0]
[0 1]
Does this not count as an inverse for some reason or is it just that there is no process to find it in general?

[peaceofheart]

After knowing this result, I m a bit confused about the validity of "inverse of singular matrices does not exist "plzz help me

[jirkakalecky]

Did you have to teach yourself how to write from right to left, just to make these videos, or is there some trick I dont get?

[MrNiceOne]

maybe slight nitpicking but, you only showed how identity might look and didn't actually prove that one you got really is identity