Very nice derivation and a superb explanation (as usual by this presentor) ! I did verified (proved ?) the formula by using your X matrix, expressing matrix A as X^2, computing its trace and det, then using the formula, showing that indeed it arrives at matrix X. But I wondered how one arrived, in the first place, at the formula - and now I know... - Thanks !
You missed a trick. Use the Cayley-Hamiltonian theorem again, X^2-tr(X)X+det(X)I_2=0. Note that taking the trace is a LINEAR operation. Take the trace to obtain: tr(X^2)-(tr(X))^2+2det(X)=0. Note that X^2=A, and det(X)=sqrt(det(A)) and rearrange to get: (tr(X))^2=tr(A)+2sqrt(det(A)), take square roots to get tr(X)=sqrt(tr(A)+2sqrt(det(A))). I think that this is slicker.
Some 2x2 matrices can have infinitely many square roots, not just up to 4. For example, a matrix of the form [[1,0],[x,-1]] is a square root of the 2x2 identity matrix for any complex number x.
There may be infinitely many solutions. For example every involution matrix (A^2=Id) is a square root if the identity matrix. But if we consider solutions up to similarity we have at most 4 solutions.
You should have precised that "the" square root of a matrix is not unique, it distrubed me in the other video. Nice tricks otherwise and very well explained ! I knew the characteristic polynom and Cayley-Hamilton but I would never have found out it by myself. Great job, very interesting !
I have another idea about the proof. Let √A = k ( A + p I ) Then A = k^2 ( A + p I )^2 = k^2 (A^2 + 2pA + p^2 I) Then use Cayley-Hamilton theorem to reduce A^2 in terms of A and I, and then comparing the coefficients of A and I on both sides and then solve for k and p.
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
Probably no, because for larger matrices the characteristic equation becomes more complicated and hence using Cayley-Hamilton like it was used here for deriving the formula does not work anymore.
you can sometimes take the square root of a singular matrix! funnily enough, I made a video about this same topic on my channel just a couple of weeks ago, but basically, if the determinant = 0, it sometimes just reduces the potential number of square roots to 2 as opposed to 4.
He should have done that with a 3x3 matrix of X = [(a,b,c), (d,e,f), (g,h,i)] solution and in terms of a + e + i = tr(X) and det(X)= aei + bfg + cdh - gec - hfa - idb = det(X). We know the eigenvalue Caly Hamilton formula is true for nth eigenvector of [A] - (lambda)[I] = 0 in studying n eigenvalues for NxN matrix [A].😑🙄🤯 I haven't proved all that but for 2x2 matrices his solution is proof enough. 😁👍
Correct my Cayly Hamilton eigenvalues idea to det[A -(lambdas)[I]] = 0. Not the A - (lambda matrix)[I] = 0 incorrect Cayly Hamilton Eigenvector stated formula. Also [(a, b, c), (d, e, f), (g, h, i)]^2 = a(a + d + g), b(b + e + h), c(c + f + i), etc for 3x3 matrix of nine calculated elements in each row and column will need to tie in with the previous three equations.
It does not extend directly to higher dimensions, because for larger matrices, the characteristic equation and hence the Cayley-Hamilton equation become more complicated: You do not only need the determinant and the trace, but also other (symmetric) combinations of the matrix elements.
@@bjornfeuerbacher5514 ... It is a theorem to extend to any NxN system of equations. It solves by the PAP-1 matrices for powers of matrix N. For example, knowing matrix A then powers are found by det| [A] - [lamda][I]| = 0 where the lamda matrix is NxN or a system of N equations for N unknowns of the N eigenvalues of an NxN State Equation matrix A. This used mostly for solving First Order linear Differential Equations in multiple N number of State Equation variables. Electrical Engineering students need this in all RLC circuit problems with circuits containing inductors, resistors and capacitors with every closed circuit loop of course V=LdI/dt, V=IR and I=CdV/dt calculations of N circuit loops and V is an N voltage variables Nx1 matrix and I is N current variables Nx1 matrix.
Probably yes. You could use the Cayley-Hamilton theorem to simplify larger powers of X to a linear expression, and use that to derive formulas for larger roots.
Tried to do it... apparently, it doesn't work. E. g. for X³ = A, one can multiply the CH equation with X in order to arrive at an equation which contains X³. But then one also has a term containing X², and one needs the trace of that - and I don't see how one could simplify that using only the trace and determinant of A. :/
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 😢
Much too complicated. Let's do it differently: Let A as described in the video. Then, let X^2 = A, with X = matrix of x1...x4. Then, a1 = x1^2 + x2*x3; a2 = x1*x2 + x2*x4; a3 = x1*x3 + x2*x4; a4 = x2*x3 + x4^2. Four equations, four unknowns. Solve. Done.
Note that tr(A) and dat(A) are invariants of the matrix. So I suspect that there is a topological derivation of this result which is quite simple in application.
My maths degree has several decades of dust on it, so forgive a perhaps silly question... For the determinant of A, if we consider the positive square root in the numerator, must we be consistent and also use the positive square root in the denominator? And similarly for the negative square root, thus leading to up to two distinct solutions? Or can we mix their signs, thus leading to up to four distinct solutions? And, I just realised, if we consider the entire denominator also can be positive or negative, up to eight distinct solutions?
Your math degree let you lose focus on [(a, b), (c, d)] matrix [X]! I think all ✓s are principle square roots or all positive because variables a, b, c and d are not stated as being negative or positive number replacement variables.
I have to clarify since we notice we'll only consider only two and only two things, the ad > bc or ad < bc cases that substituting a, b, c and d values into those variables. Also notice a square roots when ad < bc is taking the square roots of a negative. The condition ad - bc has to equal or be greater than 0 or else we have a complex number in the formula which is not good in matrices of real numbers math.
Maybe I'm rushing too much but ad can be greater or less than be or 0 in matrix X but it is true that det[a] has to positive when he takes the square roots in the formula. Substituting the a, b, c and d values for matrix A forces it's ad > bc elements condition to be true or ✓negative in determinant is a complex number calculation while if it was positive then the elements of X matrix is only for two cases since we are squaring. ad > bc and ad < bc are both okay for elements in matrix X because they'll multiply (-)(-) or (+)(+) as positive. So in summary whatever you take ✓detA as you have to be consistent in the denominator or you'll form a negative detA which is not a real number solution.
OK, answering my own question about whether the formula works for all four combinations for the signs of the formula's two square roots of det(A) What better way than to try them all and see what happens... +/+ (both positive) works as expected, -/- (both negative) also works. But neither varying of the signs... +/- nor -/+ work, though the former (when squared) gives a multiple (9) of A, and the latter (when squared) a multiple (2) of I.
@Karlston ... Answering your possible "overthink" the determinant of the squared matrix has to be positive only. Otherwise by the formula the square root of th determinant of matrix A in the derived formula goes imaginary. This teacher forgot to mention that the determinant of matrix A must be positive. That is requirement #1. Now [A] = [X][X] so we haven't figured what the determinant of [X] is figured out to be. You haven't taken an Abstract Algebra course in university math studies. So until you do you'll continue to accept a wrong matrix relationship: [A][B] = [B][A]. Matrices do not form an "abelian group!" So like all "non abelian" products only unique matrices [X] times [X] have only a unique set of elements a, b, c and d that form the matrix A elements a, b, c and d of that mateix
Huh? In the formula given here, there is no matrix in the denominator. And you _can_ do somethimg like divide by a matrix - simply multiply by its inverse matrix.
@@bjornfeuerbacher5514 What you said is correct but I wouldn't call it "division" for two reasons: 1) Matrix multiplication does not commute (so you would have a "division from the left" and another "division from the right"), and 2) the inverse doesn't always exist (only for regular matrices). In short: Square matrices do not form a field, only a (non-commutative) ring, and I would reserve the term "division" for fields.
Very poor. Math is all about meaning (and in defining concepts meticulously), and you fail to convey meaning and define things clearly. Sad for the non-mathematician people who watch this and think they’re “learning math”…
Perhaps you're in the wrong class. There's always one who will complain about something free. Maybe you don't have the prerequisites for this topic but I thought it was a excellent review. Go sit in the corner.
![[MV] A leap of faith...ความเชื่อใจครั้งสุดท้าย (From "Petrichor the series")](http://i.ytimg.com/vi/U1piZH2CNXA/mqdefault.jpg)
Congrats, You are a extraordinary professor. I became a fan! It worth watching your videos.
Very nice derivation and a superb explanation (as usual by this presentor) ! I did verified (proved ?) the formula by using your X matrix, expressing matrix A as X^2, computing its trace and det, then using the formula, showing that indeed it arrives at matrix X. But I wondered how one arrived, in the first place, at the formula - and now I know... - Thanks !
You missed a trick. Use the Cayley-Hamiltonian theorem again, X^2-tr(X)X+det(X)I_2=0. Note that taking the trace is a LINEAR operation. Take the trace to obtain:
tr(X^2)-(tr(X))^2+2det(X)=0. Note that X^2=A, and det(X)=sqrt(det(A)) and rearrange to get: (tr(X))^2=tr(A)+2sqrt(det(A)), take square roots to get tr(X)=sqrt(tr(A)+2sqrt(det(A))).
I think that this is slicker.
Some 2x2 matrices can have infinitely many square roots, not just up to 4.
For example, a matrix of the form [[1,0],[x,-1]] is a square root of the 2x2 identity matrix for any complex number x.
That's the case ad - bc = -1 - 0 = -1 and Trace is 0?
What would be a good linear algebra book for self study that has the Cayley-Hamiltonian and problems such as finding square roots of matrixes?
I wish you we around 10 years ago when I was first tackling these kinds of problems!
Superb! For linear algebra I recommend Prime Newtons.
There may be infinitely many solutions. For example every involution matrix (A^2=Id) is a square root if the identity matrix. But if we consider solutions up to similarity we have at most 4 solutions.
Excellent Job!
Very good professor!
Thanks!
You should have precised that "the" square root of a matrix is not unique, it distrubed me in the other video.
Nice tricks otherwise and very well explained ! I knew the characteristic polynom and Cayley-Hamilton but I would never have found out it by myself.
Great job, very interesting !
I have another idea about the proof.
Let √A = k ( A + p I )
Then
A = k^2 ( A + p I )^2 = k^2 (A^2 + 2pA + p^2 I)
Then use Cayley-Hamilton theorem to reduce A^2 in terms of A and I, and then comparing the coefficients of A and I on both sides and then solve for k and p.
Very interesting video using Cayley-Hamilton thr.
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
Does this formula work for matrices larger than 2x2 ?
Probably no, because for larger matrices the characteristic equation becomes more complicated and hence using Cayley-Hamilton like it was used here for deriving the formula does not work anymore.
6:02 Cayley-Hamilton theorem that is what is not present in your algebra series
but what I suggested in my comments to record video about it
It is true for 2x2 matrices but more useful for considering NxN matrix of n eigenvalues not just 2 eigenvalues!
Dressed up really nicely
Very sophisticated
Nice exercise!
You are so cool at math💚
What a nice jacket! 😄
It's apparent that A can't be singular for this to work, right? You can't have a square root of a singular matrix, is that right?
you can sometimes take the square root of a singular matrix! funnily enough, I made a video about this same topic on my channel just a couple of weeks ago, but basically, if the determinant = 0, it sometimes just reduces the potential number of square roots to 2 as opposed to 4.
Please do more MIT integration bee problems 🙏🙏
Nice! Better than rice!
Well said stop learning stop living
Nothing is better than rice, except this formula probabely!
Can this be extended to higher dimensions or is only valid for 2x2 matrices?
He should have done that with a 3x3 matrix of X = [(a,b,c), (d,e,f), (g,h,i)] solution and in terms of a + e + i = tr(X) and det(X)= aei + bfg + cdh - gec - hfa - idb = det(X). We know the eigenvalue Caly Hamilton formula is true for nth eigenvector of [A] - (lambda)[I] = 0 in studying n eigenvalues for NxN matrix [A].😑🙄🤯
I haven't proved all that but for 2x2 matrices his solution is proof enough. 😁👍
Correct my Cayly Hamilton eigenvalues idea to det[A -(lambdas)[I]] = 0. Not the A - (lambda matrix)[I] = 0 incorrect Cayly Hamilton Eigenvector stated formula.
Also [(a, b, c), (d, e, f), (g, h, i)]^2 = a(a + d + g), b(b + e + h), c(c + f + i), etc for 3x3 matrix of nine calculated elements in each row and column will need to tie in with the previous three equations.
It does not extend directly to higher dimensions, because for larger matrices, the characteristic equation and hence the Cayley-Hamilton equation become more complicated: You do not only need the determinant and the trace, but also other (symmetric) combinations of the matrix elements.
@@bjornfeuerbacher5514 ... It is a theorem to extend to any NxN system of equations. It solves by the PAP-1 matrices for powers of matrix N. For example, knowing matrix A then powers are found by det| [A] - [lamda][I]| = 0 where the lamda matrix is NxN or a system of N equations for N unknowns of the N eigenvalues of an NxN State Equation matrix A.
This used mostly for solving First Order linear Differential Equations in multiple N number of State Equation variables. Electrical Engineering students need this in all RLC circuit problems with circuits containing inductors, resistors and capacitors with every closed circuit loop of course V=LdI/dt, V=IR and I=CdV/dt calculations of N circuit loops and V is an N voltage variables Nx1 matrix and I is N current variables Nx1 matrix.
plzzz prove Cayley-Hamilton theorem
Well done
Thank you so much. God bless you 🙏❤
You're a man of your word. Thank you for the likes 😊
Is there a formula for the nth root of the 2 x 2 matrix A?
Probably yes. You could use the Cayley-Hamilton theorem to simplify larger powers of X to a linear expression, and use that to derive formulas for larger roots.
Tried to do it... apparently, it doesn't work. E. g. for X³ = A, one can multiply the CH equation with X in order to arrive at an equation which contains X³. But then one also has a term containing X², and one needs the trace of that - and I don't see how one could simplify that using only the trace and determinant of A. :/
Prime newtons you sound like Richard Mofe Damijo and I imagine you are a Nigerian
Who give this formula?
Originally it was published by Levinger, B. W. (1980). The Square Root of a 2 × 2 Matrix. Mathematics Magazine, 53(4), 222-224.
I sent question in ur mail but no response yet from 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 😢
Much too complicated. Let's do it differently:
Let A as described in the video. Then, let X^2 = A, with X = matrix of x1...x4. Then, a1 = x1^2 + x2*x3; a2 = x1*x2 + x2*x4; a3 = x1*x3 + x2*x4; a4 = x2*x3 + x4^2.
Four equations, four unknowns. Solve. Done.
I would say your version is much _more_ complicated. The version in the video was quite elegant.
Note that tr(A) and dat(A) are invariants of the matrix. So I suspect that there is a topological derivation of this result which is quite simple in application.
X=Sqrt[A] X^2=A
My maths degree has several decades of dust on it, so forgive a perhaps silly question...
For the determinant of A, if we consider the positive square root in the numerator, must we be consistent and also use the positive square root in the denominator? And similarly for the negative square root, thus leading to up to two distinct solutions?
Or can we mix their signs, thus leading to up to four distinct solutions?
And, I just realised, if we consider the entire denominator also can be positive or negative, up to eight distinct solutions?
Your math degree let you lose focus on [(a, b), (c, d)] matrix [X]! I think all ✓s are principle square roots or all positive because variables a, b, c and d are not stated as being negative or positive number replacement variables.
I have to clarify since we notice we'll only consider only two and only two things, the ad > bc or ad < bc cases that substituting a, b, c and d values into those variables. Also notice a square roots when ad < bc is taking the square roots of a negative. The condition ad - bc has to equal or be greater than 0 or else we have a complex number in the formula which is not good in matrices of real numbers math.
Maybe I'm rushing too much but ad can be greater or less than be or 0 in matrix X but it is true that det[a] has to positive when he takes the square roots in the formula. Substituting the a, b, c and d values for matrix A forces it's ad > bc elements condition to be true or ✓negative in determinant is a complex number calculation while if it was positive then the elements of X matrix is only for two cases since we are squaring. ad > bc and ad < bc are both okay for elements in matrix X because they'll multiply (-)(-) or (+)(+) as positive. So in summary whatever you take ✓detA as you have to be consistent in the denominator or you'll form a negative detA which is not a real number solution.
OK, answering my own question about whether the formula works for all four combinations for the signs of the formula's two square roots of det(A) What better way than to try them all and see what happens...
+/+ (both positive) works as expected, -/- (both negative) also works. But neither varying of the signs... +/- nor -/+ work, though the former (when squared) gives a multiple (9) of A, and the latter (when squared) a multiple (2) of I.
@Karlston ... Answering your possible "overthink" the determinant of the squared matrix has to be positive only. Otherwise by the formula the square root of th determinant of matrix A in the derived formula goes imaginary. This teacher forgot to mention that the determinant of matrix A must be positive. That is requirement #1. Now [A] = [X][X] so we haven't figured what the determinant of [X] is figured out to be.
You haven't taken an Abstract Algebra course in university math studies. So until you do you'll continue to accept a wrong matrix relationship: [A][B] = [B][A]. Matrices do not form an "abelian group!"
So like all "non abelian" products only unique matrices [X] times [X] have only a unique set of elements a, b, c and d that form the matrix A elements a, b, c and d of that mateix
(a-λ)(d-λ)-bc=0 ad-aλ-d λ+ λ^2-bc=0 λ^2-(a+d)λ+ad-bc=0 a+d=tr(X)
Cool
Amazing
tr(A)=a^2+d^2+2bc=(a+d)^2-2ad+2bc=(a+d)^2-2(ad-bc) tr(A)=(tr(X))^2-2 detX tr(X)=Sqrt[tr(A)+2det(X)]=Sqrt[tr(A)+2Sqrt[det A]] Sqrt[A]=(A+Sqrt[det(A)] I]/(Sqrt[tr(A)+2Sqrt[det A])
X= {{a,b},{c,d}} det(X-λI)=0 det({{a-λ,b},{c,d-λ}})=0
WHY THE * IS THE MATRIX IN THE FRACTIONS!? MATRICES ARE UNABLE TO DIVIDE.
Huh? In the formula given here, there is no matrix in the denominator.
And you _can_ do somethimg like divide by a matrix - simply multiply by its inverse matrix.
@@bjornfeuerbacher5514 What you said is correct but I wouldn't call it "division" for two reasons: 1) Matrix multiplication does not commute (so you would have a "division from the left" and another "division from the right"), and 2) the inverse doesn't always exist (only for regular matrices). In short: Square matrices do not form a field, only a (non-commutative) ring, and I would reserve the term "division" for fields.
@@Grecks75 Good points. :)
Very poor. Math is all about meaning (and in defining concepts meticulously), and you fail to convey meaning and define things clearly. Sad for the non-mathematician people who watch this and think they’re “learning math”…
Perhaps you're in the wrong class.
There's always one who will complain about something free. Maybe you don't have the prerequisites for this topic but I thought it was a excellent review. Go sit in the corner.
Muy buenos videos amigo, saludos