Here's a note on where matrices can take you, once you get used to using the subscript notation rather than writing them out in full. AB in general does not equal BA, or as mathematicians and physicists say matrix multiplication is "non commutative". But (AB)C always equals A(BC) (associative) And A(B+C) always equals AB + BC (distributive) In Quantum Mechanics the non commutativity combined with the other two properties makes matrix algebra very useful. The difference AB - BA in particular turns out to have important physical meaning, sometimes called the commutator, and that in turn means that in QM the important matrices are square (otherwise that difference is undefined) It is also important for matrices to be useful to QM that associativity and distributivity do work as for scalars. The slight oddity to the incoming physicist is that some QM matrices have an infinite number of rows and columns... you don't usually meet that possibility in a maths course on matrices. When writing the equations with subscripts is actually no harder than a 2x2, just don't get stressed by the fact that you are covering infinities of elements. And if that feels strange for now, maybe come back to it later when you have had more experience with sensible sized matrices! Hint: don't try to write the elements out in full though...
Can I just prove property 4 by doing this? As dimension of A is m×n and n×p for B. Hence the dimension of the product should be m×p For (AB)^T, the dimension of their product should be p×m this time and the only way to get this, is p×n matrix times n×m martix, which are B^T times A^T
I got a test with an integration question, integrate the function ax²+bc+c. Everything was well and good and I got ax³/3+bx²/2+cx+constant. *BUT* They gave one option as ax³/3+bx²/2+cx+c (i.e; instead of constant they wrote c). But isn't that wrong. Because they have chosen a variable c for constant which is already present in the equation. I lost totally 5 marks due to this..
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.
Hello sir, but I think that at 0:24, you said 2X4 matrix, but you wrote down a 4X2 Matrx which was transposed to a 4X2. I think you were thinking of the transposed, not the original.
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.
Could you define a partial transpose operation P(n) such that A^P(0)=A, A^P(1)=A^T, A^P(2)=A, and A^P(1/2)^P(1/2)=A^T? What would A^P(1/2) look like? I kind of expect it to require something like imaginary matrices, kind like how complex numbers provide continuous behavior between positive and negative numbers.
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
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
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??
Dear blackpen redpen: Can you do a 1 hour CRASH course of elementary functions such as z(x,y)= e^xy2ln+ln(3y+x) find the derivative of that, and stuff like Change of X and change of Y gives this and this, how to work with the definition of the derivative. And stuff like Quick tips for basic optimizations, some basic integral rules and ways to do it, just an overall guide for math noobs
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.
(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
@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
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Good morning!
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
4:46, oh I love that look of a Mathematician.
I just saw this today in university. Thanks, you are the best man!!!
Here's a note on where matrices can take you, once you get used to using the subscript notation rather than writing them out in full.
AB in general does not equal BA, or as mathematicians and physicists say matrix multiplication is "non commutative".
But (AB)C always equals A(BC) (associative)
And
A(B+C) always equals AB + BC (distributive)
In Quantum Mechanics the non commutativity combined with the other two properties makes matrix algebra very useful.
The difference AB - BA in particular turns out to have important physical meaning, sometimes called the commutator, and that in turn means that in QM the important matrices are square (otherwise that difference is undefined)
It is also important for matrices to be useful to QM that associativity and distributivity do work as for scalars.
The slight oddity to the incoming physicist is that some QM matrices have an infinite number of rows and columns... you don't usually meet that possibility in a maths course on matrices. When writing the equations with subscripts is actually no harder than a 2x2, just don't get stressed by the fact that you are covering infinities of elements. And if that feels strange for now, maybe come back to it later when you have had more experience with sensible sized matrices!
Hint: don't try to write the elements out in full though...
Heyyy are u math teacher??
This came at the right time tq mr blackpenredpen !!!!!
No dual spaces? 😭😭😭
I don’t even have a space in the first place
Thank You very much!
I want this in my subject in linear Algebra in early 9th grade.🙏😱
that last prove was a lifesaver dude, thanks
This was so helpful and well-explained, thank you so much!
You r best 👍👍👍 maths teacher. And i want to be like u 😍😍😋 lots of love from india 🇮🇳🇮🇳
Can I just prove property 4 by doing this? As dimension of A is m×n and n×p for B. Hence the dimension of the product should be m×p
For (AB)^T, the dimension of their product should be p×m this time and the only way to get this, is p×n matrix times n×m martix, which are B^T times A^T
Just started transposition today .....And this will act as revision xd .
Btw my man has grown a beard!!!
4:46 In general, AT.BT does not exist because the dimensions are incompatible. AT is n x m and BT is p x n. Is that correct?
'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.
Love from India ❤❤
He is Mathbulious!!! 🔥🔥
I am excited because in this year I learn matrices and limits
Hey bprp! I have a question for you!
Can you PINPOINT the value of 100^2 upto 10 decimals with or without using calculators?
thank you so much you saved me
So easy. Can do it over any field. Or even a ring, where a matrix is invertible iff its det is a unit.
When you multiply matrices ...it's weird.'
---blackpenredpen
If a matrix represents a linear mapping, then what mapping does its transpose represent?
Check out my dual space playlist for an answer to this ;)
1 and 4 properties are nearly the same for matrix inverse (but for matrix inverse we must have square and non singular matrices)
"It's kinda wierd because when you multiply matrices, it's wierd"
nice job
I got a test with an integration question, integrate the function ax²+bc+c. Everything was well and good and I got ax³/3+bx²/2+cx+constant.
*BUT*
They gave one option as ax³/3+bx²/2+cx+c (i.e; instead of constant they wrote c). But isn't that wrong. Because they have chosen a variable c for constant which is already present in the equation. I lost totally 5 marks due to this..
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.
Hello sir, but I think that at 0:24, you said 2X4 matrix, but you wrote down a 4X2 Matrx which was transposed to a 4X2. I think you were thinking of the transposed, not the original.
Hi why is the utility of finding Transpose of a Matrix? Thank you
I have a question. Using properties 2 and 3 transposition is linear operator. My question- what the matrix of this operator?
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.
Morning Steve ☀
Can you do some fuctional equations after linear algebra rush.
Love form Nepal❤
😄👍
You are the best teacher
finally algebra seRIES!!!!!!
Could you define a partial transpose operation P(n) such that A^P(0)=A, A^P(1)=A^T, A^P(2)=A, and A^P(1/2)^P(1/2)=A^T?
What would A^P(1/2) look like?
I kind of expect it to require something like imaginary matrices, kind like how complex numbers provide continuous behavior between positive and negative numbers.
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
Good night bruh ..... We from india... Waiting for your post
3:19 me with everything
Is the condition of having the appropriate dimensions the only reason matrix multiplication is not commutative?
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.
Why do you not add sung outro anymore? 🎶Black pen red pen...
I learned to hate Linear Algebra quite immensely in my younger days.
Are properties 2 and 3 because matrix addition and scalar multiplication point-wise?
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.
(1-x^2)y'=1-xy-3x^2+2x^4 can u solve this equation
4:47 serious mode on
回憶在中學時的課程
2020 raise to the power 2019 - 2020 divided by 2020 square + 2021=N
then find the sum of digits of n
bro plz solve this?? trying from last 5 weeks
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
where did 7:03 come from?
In other words, the matrix transpose is a determinant-preserving linear involutive antiautomorphism on the set of matrices.
Cool!
Matrix Transpose:
Rows become columns and columns become rows
Can you solve algebra problems
Can you solve (AB)^T = A^T B^T for A and B?
If A and b commute.
@@blackpenredpen What do you mean by commute?
@@Pkaysantana AB = BA
already knew that.
9:12 , your 'not a box' still looks more box than my boxes °Π°
Pf: 2 please
thats interested
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if A and B are commutable,their transposes too.
1, 2, 3, 4, 5, 6, 7, ... 9!?!?!?!? How dare you, sir!
Dear blackpen redpen: Can you do a 1 hour CRASH course of elementary functions such as z(x,y)= e^xy2ln+ln(3y+x) find the derivative of that, and stuff like Change of X and change of Y gives this and this, how to work with the definition of the derivative. And stuff like Quick tips for basic optimizations, some basic integral rules and ways to do it, just an overall guide for math noobs
And number theory
Hermite and Smith normal forms
What's up with the beard lmao 😂
How to find the square root of matrix
I’ve done that on my channel
@@drpeyam thanks
I thought this video be about commutative matrices
oh the pokemon ball is his mic.......
en.m.wikipedia.org/wiki/Symplectic_matrix not directly a transpose property but maybe you can explain in a future episode 😁
The transpose of 8 is infinity 😅
i love you 😘😘😘😘😘❤️
Want a bprp best moments episode
This is so trivial.
We, the people of the world would like you to see with the full-touch chalk. 🔥Will you?🔥
Yes I would too!
u always hold something cute in your left hand... does it really help keeping people’s eyes on the screen?
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.
Indian teachers are best...😊😊not foreign
Lo único que entendí fue pokeball.
Pokemon ball?? 🙁
First
U look weird in that beard
Nah
@@blackpenredpen ohhk
❤️
2020 raise to the power 2019 - 2020 divided by 2020 square + 2021=N
then find the sum of digits of n
bro plz solve this?? trying from last 5 weeks
(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
First
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First