Maybe I’m driving you nuts maybe I’m driving everybody nuts but I’ve asked this 100 times. We’re in the world did you get your math expertise and do you teach anywhere. It is truly amazing both your expertise and your teaching. Thanks
Wow wonderful explanation! I never thought of square root of matrix, because our teachers never taught this thing. Your way of explanation is mind blowing! Especially Your smile is awesome. Keep shining sir
Generalize it for nxn square matrix Do we need symmetric and positive definite matrix for Choleski decomposition ? Matrix A is symmetric when A = A^{T} Matrix A is positive definite when for all non-zero vectors x x^{T}Ax > 0 Choleski decomposition A = LL^{T} but in textbooks is written that A must be symmetric and positive definite
Without much hassle, you obtain the OTHER square root by changing +√det into -√det in both the numerator and denominator. (and of course, there are two other square roots by changing the sign of B, for a total of 4 solutions.)
It is interesting to compare algebra of numbers with the algebra of matrices. Every number has two different square roots (possibly complex), except 0 has only 1. When dealing with 2 by matrices you have the following result: every 2 by 2 matrix has 4 square roots (possibly complex) or 2 square roots or 1 square root or no root or infinitely many square roots.
Here is an other solution : let B = [ (a, b) (c, d) ]. Then B^2 = A leads to : (E1) a^2 + bc = 2 (E2) b (a + d) = 1 (E3) c (a + d) = 2 (E4) d^2 + bc = 3 Then, (E4) - (E1) gives d^2 - a^2 = 1, with obvious solution d = 1 and a = 0. (E2) and (E3) give b = 1 and c = 2. So, an other solution is B = [ (0, 1) (2, 1) ]. Thank you for your interesting videos ! 🙂 NB : never heard about your magical formula… and hope you show how you get it !
I think I’ve got it for the « magical formula »… Starting from a well-known property in M2 (R) (and M2 (C)) : any matrix B checks the equation B^2 - trB . B + ∆B . I = 0, where trB is the trace of B, ∆B is the determinant and I is the unit matrix of M2 (R). So we have : trB . B = B^2 + ∆B . I or B = (1 / trB) [B^2 + ∆B . I] (and of course, B is a square root of B^2 !) Let’s take the trace left and right : trB . trB = tr (B^2) + tr (∆B . I) or (trB)^2 = tr (B^2) + 2 ∆B And we notice that ∆(B^2) = (∆B)^2. This leads to trB = √ (tr (B^2) + 2 ∆B), and B = [1 / √ (tr (B^2) + 2 √∆(B^2))] . [B^2 + √∆(B^2) . I] … which is the « magical formula »… 🙂
Hi, I'm from Mexico, and I´m studying computing engeneer, and this kind of exercises caught my attention, this formula or this topic I've never seen on my Linear Algebra course, and I would like to know how can I find this theme or if this is particularly on a Lineal Algebra Course, Very nice video i learned something new. Thanks
Have to give Prime Newtons credit: he is a master teacher. And that itself no easy or simple thing to be or even describe in a few words, but it's something like being a master musician.
I would have used the Eigen decomposition [A]^(0.5)=[Q] [/\]^(0.5) [Q]^(-1) where [Q] is the eigenvector matrix and [/\] is eigenvalue diagonal matrix. This works because raising a diagonal matrix to a power is trivial. You simply raise each diagonal element to the power🙂
You can do practically everything in math! Can you integrate absolute value, arcfunctions and fractional part? Can you show the chess horse puzzle's solution?
I'm thinking one could also define √A as its taylor series, akin to how one defines the exponential of a matrix: e^A = I + A + (1/2!)A^2 + ... Though this is way more interesting. Being in closed form is also an advantage over having to deal with an infinite series 😅
Hello, my exam just got over, I can finally watch your videos in peace! If you don't mind u have a good question of geometry about quadrilaterals, I will email it to you. Please take a look at it. It's okay if you don't make a video about it. I would be happy with just the solution. Thank you!
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
i can assume there is a 2X2 matrix which multiplied twice gives the matrix inside the sruare root. and in this way I get 4 equation with 4 variable. then solve them to find the answer
Sir what is the meaning of this ? can we calculate without formula? just like multiplication of matrix or squar, cube or nth power of matrix. please make a video.
@@budderman3rd Depends on conventions. In high school algebra it is normally just √4̅ = 2, though for complex numbers it is normally √−̅2̅𝒊̅ = ±(1−𝒊), since we have no good universally accepted method to choose between 1−𝒊 and −1+𝒊 here 🤷♂️
@@sowndolphin5386 Yes, that's the main reason (for the real-valued square root). On the other hand, √x̅ may be regarded as a _multivalued_ function which returns two opposite values for non-negative real radicands. From Wiki: "Every real number greater than zero has two real square roots, so that square root may be considered a multivalued function. For example, we may write: *√4̅ = ±2 = {2; −2}.* Each nonzero complex number has two square roots, three cube roots, and in general *n* nth roots."
But √4̅ = ±2, not just √4̅ = 2, isn't it? So we have (at least) one more solution: [(2-2), (1-0); (2-0), (3-2)]/√5̅-̅4̅ = [0, 1; 2, 1] And also these two solutions with a negative sign, which gives us four solutions in total.
No no, if u say that √4 = ±2, u are basically violating the fundamental of mathematics by saying a real number √4 has two values 2 and -2 which isn't possible, that's why sqrt function only has the range of non negative real numbers √4 is only equal to 2 sqrt(x²) = |x| (absolute value)
@@ExquisiteHappiness Your argument makes no sense in the context of the solution of the problem, since for A = [2, 1; 2, 3] and B₂ = [0, 1; 2, 1] we indeed get (B₂)² = A (you may easily verify it yourself).
@@ExquisiteHappiness > a real number √4 has two values 2 and -2 No, √4̅ is not necessarily a uniquely defined real number - it is a (multivalued or single-valued, depending on our choice) function from ℂ to ℂ (in the general case), so we can easily choose it to be √4̅ = ±2 = {+2; -2} - a two-element set of square roots, just like ³√-̅8̅ = {-2; 1+𝒊√3̅; 1-𝒊√3̅} is a three-element set of complex cube roots of -8.
@@-wx-78- Yeah, sure: that's why I said in my top comment that "And also these two solutions with a negative sign, which gives us four solutions in total". That's due to the denominator having two opposite square root values as well.
I thought all of these guys only watch the videos and live streams from a young man, who records ’em all in his basement in Herne, Northrhine-Westphalia, Germany.
Cannot wait for the formula's proof video! Please make it the next upload!!!
I lived long enough to see the square root of a matrix. Magical, and in hindsight, it makes perfect sense.
Where were you when you heard about Nixon’s Watergate scandal?
Your videos really help me with my JEE preparation, thank you so much sir !!
so true , some of his videos concepts are used in questions of jee mains and advance
Maybe I’m driving you nuts maybe I’m driving everybody nuts but I’ve asked this 100 times. We’re in the world did you get your math expertise and do you teach anywhere. It is truly amazing both your expertise and your teaching. Thanks
Wow wonderful explanation! I never thought of square root of matrix, because our teachers never taught this thing. Your way of explanation is mind blowing! Especially Your smile is awesome. Keep shining sir
Generalize it for nxn square matrix
Do we need symmetric and positive definite matrix for Choleski decomposition ?
Matrix A is symmetric when A = A^{T}
Matrix A is positive definite when
for all non-zero vectors x
x^{T}Ax > 0
Choleski decomposition A = LL^{T}
but in textbooks is written that A must be symmetric and positive definite
Excellent problem and solution. Good choice of numbers too.
I wish that I had you as a teacher.
Without much hassle, you obtain the OTHER square root by changing +√det into -√det in both the numerator and denominator.
(and of course, there are two other square roots by changing the sign of B, for a total of 4 solutions.)
I am really getting better every day because of your videos☺️☺️
It is interesting to compare algebra of numbers with the algebra of matrices. Every number has two different square roots (possibly complex), except 0 has only 1. When dealing with 2 by matrices you have the following result: every 2 by 2 matrix has 4 square roots (possibly complex) or 2 square roots or 1 square root or no root or infinitely many square roots.
Thanks! I like to see the proofs in order to understand where the formula came from.
Here is an other solution : let B = [ (a, b) (c, d) ]. Then B^2 = A leads to :
(E1) a^2 + bc = 2
(E2) b (a + d) = 1
(E3) c (a + d) = 2
(E4) d^2 + bc = 3
Then, (E4) - (E1) gives d^2 - a^2 = 1, with obvious solution d = 1 and a = 0.
(E2) and (E3) give b = 1 and c = 2.
So, an other solution is B = [ (0, 1) (2, 1) ].
Thank you for your interesting videos ! 🙂
NB : never heard about your magical formula… and hope you show how you get it !
We can notice that if B is a solution, the opposite - B is also a solution. We then have at least four square roots for A...
I think I’ve got it for the « magical formula »…
Starting from a well-known property in M2 (R) (and M2 (C)) : any matrix B checks the equation
B^2 - trB . B + ∆B . I = 0, where trB is the trace of B, ∆B is the determinant and I is the unit matrix of M2 (R).
So we have : trB . B = B^2 + ∆B . I or B = (1 / trB) [B^2 + ∆B . I] (and of course, B is a square root of B^2 !)
Let’s take the trace left and right : trB . trB = tr (B^2) + tr (∆B . I) or (trB)^2 = tr (B^2) + 2 ∆B
And we notice that ∆(B^2) = (∆B)^2.
This leads to trB = √ (tr (B^2) + 2 ∆B), and B = [1 / √ (tr (B^2) + 2 √∆(B^2))] . [B^2 + √∆(B^2) . I]
… which is the « magical formula »… 🙂
If u watch his latest video he mention it
i have a math degree and i've never heard about this before! so beautiful
Hi, I'm from Mexico, and I´m studying computing engeneer, and this kind of exercises caught my attention, this formula or this topic I've never seen on my Linear Algebra course, and I would like to know how can I find this theme or if this is particularly on a Lineal Algebra Course, Very nice video i learned something new. Thanks
Have to give Prime Newtons credit: he is a master teacher. And that itself no easy or simple thing to be or even describe in a few words, but it's something like being a master musician.
I would have used the Eigen decomposition [A]^(0.5)=[Q] [/\]^(0.5) [Q]^(-1) where [Q] is the eigenvector matrix and [/\] is eigenvalue diagonal matrix. This works because raising a diagonal matrix to a power is trivial. You simply raise each diagonal element to the power🙂
You can do practically everything in math! Can you integrate absolute value, arcfunctions and fractional part? Can you show the chess horse puzzle's solution?
I'm thinking one could also define √A as its taylor series, akin to how one defines the exponential of a matrix: e^A = I + A + (1/2!)A^2 + ... Though this is way more interesting. Being in closed form is also an advantage over having to deal with an infinite series 😅
Hello, my exam just got over, I can finally watch your videos in peace! If you don't mind u have a good question of geometry about quadrilaterals, I will email it to you. Please take a look at it. It's okay if you don't make a video about it. I would be happy with just the solution.
Thank you!
@srisaishravan5512 > a good question of geometry about quadrilaterals
Is parallelogram a trapezoid? 🤔
For a llgm to be considered a trapezium,1 pair of sides must be parallel a llgm satisfies the condition so the answer is yes.
@@srisaishravan5512 Some say that *only one* pair of sides must be ll, and the other two sides must not be ll 🤷♂️
So interesting! Thanks.
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
i can assume there is a 2X2 matrix which multiplied twice gives the matrix inside the sruare root. and in this way I get 4 equation with 4 variable. then solve them to find the answer
And how many solutions are we supposed to get? 🤔
The squareroot of a matrix is a SPINOR!!!!!
{{1.3 recurring,0.3 recurring},{0.6 recurring,1.6 recurring}}^2={{2,1},{2,3}} Input interpretation
(1.3^_ (repeating decimal) | 0.3^_ (repeating decimal)
0.6^_ (repeating decimal) | 1.6^_ (repeating decimal))^2 = (2 | 1
2 | 3)
Result
True
Very nice expalation.
Loved it!
Nice vid even though I don’t really understand
I thought you were just gonna use a 4 equation system, nice theorem !
We want to know the proof of the formula!
Paolo from Italy 😍🥰
You’re well educated.
Paul Dirac also found taking the square root of a matrix to be a challenging problem.
i wonder why it isn't just 4. just solving the matrix and taking a square root...i wanna know the constraints why we can't DO THIS WAY
Sir what is the meaning of this ? can we calculate without formula? just like multiplication of matrix or squar, cube or nth power of matrix. please make a video.
can u show us how to use the 3d rotation matrix for rotation on both x and y axis
Could we get a video on change of coordinates matrices
What is this applicable for?
You "lost" me at 2:02. 😂
Mister Prime Newtons, can you please show the proof of the formula displayed in 2:51
Yikes that looks heavy. Makes me want to just say "the answer is (a b, c d)", square it, and solve the resulting implied equations
I would have gone through diagonalisation --> Eigenvalue and Eigenvector
I came up with what seems to be another solution:
0 1
2 1
(Sorry, I can’t find away to put brackets round the matrix)
Yes. Usually 4 different roots. This is correct, too.
Perfect video
Next Logarithm of a matrix and its derivation also .
Long live sir
Hello, One question: what are the hypothesis to avoid that sqrt(4)=-2?
In fact, for the _algebraic_ square root, it is √4̅ = ±2, though for the _arithmetic_ square root, a so-called _principal value_ is only √4̅ = +2
maybe they wanted the graph of sqrt(x) to be a function and they eliminated the bottom part of it
@@cyberaguaThe only correct answer is the principal root. It will never be multiple answers unless its a polynomial equation
@@budderman3rd Depends on conventions. In high school algebra it is normally just √4̅ = 2, though for complex numbers it is normally √−̅2̅𝒊̅ = ±(1−𝒊), since we have no good universally accepted method to choose between 1−𝒊 and −1+𝒊 here 🤷♂️
@@sowndolphin5386 Yes, that's the main reason (for the real-valued square root). On the other hand, √x̅ may be regarded as a _multivalued_ function which returns two opposite values for non-negative real radicands.
From Wiki: "Every real number greater than zero has two real square roots, so that square root may be considered a multivalued function. For example, we may write: *√4̅ = ±2 = {2; −2}.* Each nonzero complex number has two square roots, three cube roots, and in general *n* nth roots."
Basta svolgere i calcoli..=(4/3 1/3
2/3 5/3)
can you solve ff(x) = sin(x) 😧
any other data? what to solve?
@@santhosh_not_found if f(f(x)) = sinx whats f(x)
Why would you want to take the root of a matrix?
But √4̅ = ±2, not just √4̅ = 2, isn't it? So we have (at least) one more solution:
[(2-2), (1-0); (2-0), (3-2)]/√5̅-̅4̅ = [0, 1; 2, 1]
And also these two solutions with a negative sign, which gives us four solutions in total.
No no, if u say that √4 = ±2, u are basically violating the fundamental of mathematics by saying a real number √4 has two values 2 and -2 which isn't possible, that's why sqrt function only has the range of non negative real numbers
√4 is only equal to 2
sqrt(x²) = |x| (absolute value)
@@ExquisiteHappiness Your argument makes no sense in the context of the solution of the problem, since for A = [2, 1; 2, 3] and B₂ = [0, 1; 2, 1] we indeed get (B₂)² = A (you may easily verify it yourself).
@@ExquisiteHappiness > a real number √4 has two values 2 and -2
No, √4̅ is not necessarily a uniquely defined real number - it is a (multivalued or single-valued, depending on our choice) function from ℂ to ℂ (in the general case), so we can easily choose it to be √4̅ = ±2 = {+2; -2} - a two-element set of square roots, just like ³√-̅8̅ = {-2; 1+𝒊√3̅; 1-𝒊√3̅} is a three-element set of complex cube roots of -8.
@@cyberagua Moreover, both -B₁ and -B₂ are square roots of A too. 😉
@@-wx-78- Yeah, sure: that's why I said in my top comment that "And also these two solutions with a negative sign, which gives us four solutions in total". That's due to the denominator having two opposite square root values as well.
Anyone from India ❤
me
I thought all of these guys only watch the videos and live streams from a young man, who records ’em all in his basement in Herne, Northrhine-Westphalia, Germany.
Show the proof of the square root of a square matrix. (Show the proof of the square root of a square matrix.).
I no english , solve JordanForm, aigenvectors cases
I love you forever and ever.
Ans =2...(may be )
❤ love this