Properties of the transpose of a matrix, linear algebra tutorial 

We r having evening here!! India 🇮🇳🇮🇳

It's nearly my bedtime when I saw this, about half an hour after release

Good evening

i'm from indonesia, now in indonesia is night

@@logicalproofs7276 His ❤ is his ❤ none of your ❤. Find your own gf

Heyyy are u math teacher??

I don’t even have a space in the first place

'Only' when p is unequal to m.

Exactly. If m=p AT*BT may exist but it would be still different from BT*AT

(AB)^T=B^T*A^T

If both the matrices are square matrices of same order , then it is possible to apply the property

@@SimoneCasciaro54 Not usually equal. There is nothing to stop a special case where they do happen to be equal.

Check out my dual space playlist for an answer to this ;)

😄👍

It is a question worth asking. It turns out that it is fairly straightforward to show that the transpose operation is not capable of being performed by matrix multiplication. The formal way is by reductio ad absurdam. The crafty and unofficial way is to notice that mathematicians are usually rather parsimonious with notation. They probably would not have invented a specific transpose operator if it could be reduced to a matrix and a multiplication. But let's do the proof property Consider the two by two matrix ( a, b ) ( c, d ) Assume that there exists a matrix T that when it premultiplies A gives the transpose of A. Let T= ( p, q ) ( r, s ) The top left element of the transpose is just a. That means that for any a,c we have (p,q).(a,c) = a. ((Notice here I am using a dot product. You can imagine i am writing the second vector as a column if you prefer -- I am just mixing notations because of the restrictions that Y-T comments impose)) Clearly (p, q) = ( 1, 0 ) Likewise consider the bottom right: for that corner to work out we end up with ( r, s ) = ( 0, 1) Reassembling T we find that we have the identity matrix. That clearly is not the T we set out to find. That's the first contradiction. Had we tried the other two corners we would have found a matrix ( 0, 1 ) ( 1, 0 ) so we have found contradictory values for each of the four elements of T. The logic is impeccable, and can be done for any size matrix, so our assumption that any T exists must be false. A corollary of this result is that not all linear transformations can be expressed by matrix multiplication (at least under the definition you are using for a linear transformation) This is despite the fact that a matrix multiplication always gives a linear transformation. The relationship is not reflexive.

I dont think it is possible to express transposition as a combination of matrices you could multiply your matrix A by. The transpose function is linear but from Mnp ->Mnp, so if you wanted to express that as a matrix you would have to define a matrix T which would have a size of n*p by n*p and and have it act on vectors of length n*p which would then represent your n by p matrices. With that in mind I don't doubt for a second that someone somewhere tried to figure out half-transposition lmao

Oh woops just noticed that you werent actually multiplying P and A together my bad pal, point still stands though

@@Grilnid regarding matrix dimensions, perhaps it could be first extended to a square matrix using rows and columns from an identity matrix.

Yeah so I tried something on paper and you could express transposition as a fourth-order tensor acting on matrices (aka second-order tensors) but honestly if we're delving into tensor territory I'm just way out of my depth lmao

Yes. And similarly multiplication is different because it's more complicated than pointwise

@@trueriver1950 But it's still a neat expression because it is row to column. Kind of like the transpose operator itself.

Did you mean commutative? AB is not generally equal to BA (non commutative), though there are special cases where that holds "by coincidence".

@@trueriver1950 Oh yeah sorry I'll edit it.

If A and b commute.

@@blackpenredpen What do you mean by commute?

@@Pkaysantana AB = BA

Wow that was unfortunate... this is exactly the reason why I have trust issues with mcq , like there would be error something like this and I would constantly think was it intentional or didn't caught examiner eye

@@yashsinghal1023 Everyone in my class blindly used formula and got full marks and the teacher who teaches also blindly uses formula. It seems only I got the answer wrong.

Marks doesn't equals to knowledge right!!

@@chinni6613 but my knowledge is tested only using marks.

🧡💙💚

Hermite and Smith normal forms

Pleaz can you write your question in order and use ( )

Tony Haddad when (2020^2019 - 2020) is divided by (2020^2 + 2021) then we get the remainder as N now you have to find sum of digits of the N Understood??

Tony Haddad Answer is proper numeric

@@omshandilya8888 thank u man

@@omshandilya8888 i will try my best

I’ve done that on my channel

@@drpeyam thanks

Yes I would too!

Wesley Suen, it’s his microphone... and his trademark.

It always used to be a black sphere that (in my opinion) wasn't that cute. I used to wonder if it had remote controls for the video camera on it. Then one day his gf gave him a cute toy to use as a mic holder, and since then he has varied then from time to time. It's become his second trademark, his principal trademark being the way he swaps between two colours without putting either pen down.

Nah

@@blackpenredpen ohhk

(2020²⁰¹⁹ - 2020) / ( 2020² + 2020 + 1) x = 2020 x(x²⁰¹⁸ - 1) /( x² + x + 1) = x(x - 1)( x²⁰¹⁸-1) / ( x³ - 1) Try using generating functions : ) btw N is not an integer ig ಠ_ಠ, it leaves 4080402{in a previous version I had put 2019 here by mistake} as a remainder so doesn't make much sense to talk about sum of digits Actually polynomial reduction also works well

Hamiltonian Path on dodecahedron I am really grateful that you spend time in answering but can you plz tell ahead as I just watched a video on generating function and as told by him generating function is sum of a series and now i am not able to relate it to this Q pleases guide

ya actually we need to find sum of the digis of remainder of that division

@OMS PLAYER oh , for remainder , you should try polynomial reduction, ELABORATE METHOD: consider the eqn (x^2019 - x) = Q(x) (x^2 + x + 1) + R(x) Observe that R(x) would be linear , hence x^2019 - x = Q(x) (x^2 + x + 1) + ax + b substitute x = cbrt(1) = omega (I will represent it as w) and omega ^2 so the eqn becomes w^2019 - w = Q(w)( 0 ) + a w + b - - - - - (1) similarly w^4038 - w^2 = a w^2 + b - - - - - - - (2) (1) can be written as 1 - w = aw + b and (2) as 1 - w^2 = a w^2 + b Solving gives a = -1, b = 1 Hence x^2019 - x = Q(x) (x^2+x+1) + 1-x replacing x = 2020 , we get 2020^2019 - 2020 = Q(2020)(2020^2 + 2020 + 1) - 2019 Now adding 2020^2 + 2020 + 1 to -2019 gives the required remainder = 2020^2 + 2 = 4080402 Shortcut Method avoiding complex numbers a bit we 'll again use remainder theorem but in a cleverer way in the previous method we basically substituted x such that divisor becomes 0 ie (x^2 + x + 1) = 0 => (x^3 = 1) hence in a vague sense(actually modulo x^2 + x +1) x ^ 2019 = (x^3)^k = 1 x^2019 - x = 1 -x now the previous method follows

@@omshandilya8888 You may ask if there are any queries regarding the solution