# Condition for AB+A+B=0 where A and B are matrices

• Physics lover
In summary: N^{(A)}_{n-1}## being the nilpotent matrices of degree at most ##n-m## and with the property that ##AB=0##.
Physics lover
Homework Statement
Question-:A and B be 3×3 matrices such that AB+A+B=0 then which of the following holds-:
(a)AB=-BA (b)AB=BA (c)AB=1
Relevant Equations
Matrix properties
I first tried by assuming the matrices but it was becoming complicated so i tried taking transpose on both sides,it also did not help.So now i could not think of what to do further.Help please.

With the usual commutator definition ##[X,Y]=XY-YX## try to calculate ##[A,B]=[AB-B,AB-A]=\ldots\,## If I made no sign error, this will lead to an answer.

fresh_42 said:
With the usual commutator definition ##[X,Y]=XY-YX## try to calculate ##[A,B]=[AB-B,AB-A]=\ldots\,## If I made no sign error, this will lead to an answer.
Can you please explain in simple words.I can not understand the commutator definition you are talking about.

The commutator which I defined is a bilinear, antisymmetric operation: ##[X,Y]=-[Y,X]##.
Here it is just a short way to calculate with products. If you like, you can calculate with ##XY-YX## instead, but this is confusing long as you will need twice as much terms.

Since ##A=AB-B## and ##B=AB-A##, we can calculate ##[A,B]=[AB-B,AB-A]= \ldots ## and use bilinearity and antisymmetry. Just do it. It is too easy and against the rules to do it for you.

fresh_42 said:
The commutator which I defined is a bilinear, antisymmetric operation: ##[X,Y]=-[Y,X]##.
Here it is just a short way to calculate with products. If you like, you can calculate with ##XY-YX## instead, but this is confusing long as you will need twice as much terms.

Since ##A=AB-B## and ##B=AB-A##, we can calculate ##[A,B]=[AB-B,AB-A]= \ldots ## and use bilinearity and antisymmetry. Just do it. It is too easy and against the rules to do it for you.
Can you please solve by any other method as i have not learned the commutator matrix in my school.

##(AB-A)(AB-B)-(AB-B)(AB-A) = \ldots ## compared to ##(AB-B)(AB-A)-(AB-A)(AB-B) = \ldots ##

Edit: By the way, the first link I got on a Google search for "commuting matrices" gave me https://en.wikipedia.org/wiki/Commuting_matrices where the second line is the commutator definition. You can easily say, that you looked it up on Wikipedia.

fresh_42 said:
##(AB-A)(AB-B)-(AB-B)(AB-A) = \ldots ## compared to ##(AB-B)(AB-A)-(AB-A)(AB-B) = \ldots ##

Edit: By the way, the first link I got on a Google search for "commuting matrices" gave me https://en.wikipedia.org/wiki/Commuting_matrices where the second line is the commutator definition. You can easily say, that you looked it up on Wikipedia.
Ok sir i will se that but i want to know is there any other method to solve this one.

If you perform the two calculations you might see whether it can be done with less terms. I just did a quick calculation which gave me a result - IF I didn't make any sign errors, which can easily occur in such a computation.

Answer (c) can easily be ruled out by a counterexample. So what is left is the question whether ##[A,B]=0## (b) or ##[A,B]=2AB## (a). Hence it is quite natural to compute ##[A,B]## and see where it leads to. Now all we have is ##A=AB-B## and ##B=AB-A## and thus the question naturally ends up in ##[AB-B,AB-A] \in \{\,0,2AB\,\}## and which one is true.

Maybe you can operate with the trace, but I don't have the trace formulas in mind. The determinante won't work because we have an addition as condition. So if you would have had mentioned the trace under relevant equations, I might have sought for another solution, but you didn't, and I explained why ##[A,B]## is the natural way to go.

Physics lover said:
Ok sir i will se that but i want to know is there any other method to solve this one.
my take:
selecting ##A=0## and ##B=0## obviously satisfies the equation which knocks out choice (iii)

now consider a few basic things:
##AB = -(A+B)##
##(AB)^2 = (-1)^2(A+B)^2 = A^2 + AB +BA +B^2##
simplify for case (i) and case (ii)

if you go back to the original problem and square it you get
##\big(AB +A+B\big)^2 = \mathbf 0##
so
##(AB)^2 +A^2+B^2 +ABA + AB^2 +A^2B+ BAB + AB +BA= \mathbf 0##
##\big\{(AB)^2 +(A^2+B^2)\big\} +(ABA + A^2B)+ (AB^2 + BAB) + (AB +BA)= \mathbf 0##
simplify that.

in case (i) it should imply nilpotence of ##(AB)## and in case (ii) it should imply idempotence of ##(-AB)## -- which in combination with commuting matrices means simulataneously diagonalizable triangularizable (and nice trace properties with rank).

the intersection of (i) and (ii) then is when ##AB## is the zero matrix because otherwise nilpotent matrices are not diagonalizable.

I still don't have a particularly clean finish, but you might try solving this by assuming ##A## and ##B## are diagonal matrices and seeing if any complex numbers fit this bill. Scalar field was not stated nor were any relevant equations given so I'm guessing at what is useable, though ideas related to commuting and working in ##\mathbb C## seems reasonable to get going.

- - - -
edit:
if you work through case (ii) of commuting, carefully, what you get is a nice blocked structure, after similarity transform
where ##n=3 ##
##m \in\{0,1,..n\}##

 ##(AB) = \begin{bmatrix} -I_m & \mathbf 0 & \\ \mathbf 0& \mathbf {00}^T_{n-m} \end{bmatrix} = A \cdot B = \begin{bmatrix} J_m & \mathbf 0 & \\ \mathbf 0& N^{(A)}_{n-m} \end{bmatrix} \begin{bmatrix} -J_m^{-1} & \mathbf 0 & \\ \mathbf 0& N^{(B)}_{n-m} \end{bmatrix}##
with ##N^{(B)}_{n-m}, N^{(A)}_{n-m}## being nilpotent and commuting and obeying
##N^{(B)}_{n-m}N^{(A)}_{n-m} = \mathbf {00}^T_{n-m}##

and the matrices ##J_m## having non-zero eigenvalues.

This is reducible in blocked structure and if we ignore the nilpotent piece, we see that that ##B## is in effect the negative inverse of ##A## (sometimes called a pseudo inverse)

.. I of course agree with Fresh that this problem seems to be designed to look up the definition of a commutator and applying it, though OP seems quite against doing so -- I dropped in some other ideas accordingly.

also there's a link with Fibonnaci numbers in here...

Last edited:
Using the way with the commutators, in abbreviated form ##[A,B]## or explicitly ##AB-BA## does the job. No additional assumptions necessary, especially as we do not have any information about the field and its characteristic! However, characteristic ##2## has to be excluded. This can be done as (a) and (b) would be the same in this case. But nilpotency or diagonalizability are irrelevant.

fresh_42 said:
Using the way with the commutators, in abbreviated form ##[A,B]## or explicitly ##AB-BA## does the job.
Yes, but the OP has stated that he/she hasn't learned about commutators. In any case, commutators aren't presented in an intro linear algebra course, at least that was true the last time I taught such a course about 20 years back.
fresh_42 said:
No additional assumptions necessary, especially as we do not have any information about the field and its characteristic! However, characteristic ##2## has to be excluded. This can be done as (a) and (b) would be the same in this case. But nilpotency or diagonalizability are irrelevant.

Yes, but a) I explained that he could have gotten the hint by a Google search + Wikipedia, b) that it is the natural way to do, since the question is accordingly, and c) that there is no need to calculate ##[A,B]## because it could as well be done with ##AB-BA##, which again comes naturally by the suggested answers.

When tutoring school kids, I frequently get to the problem of how to solve a linear equation system with two variables and two equations. They teach three procedures here, which are called: addition method, substitution method, and equality setting method. I find this stupid. It could as well easily done with a matrix, especially in the two by two case. I would understand it, if they learned to inspect the system and decide which method fits best, but those kids who can, can also learn the matrix method. And the others are happy if I tell them that substitution always works. I'm no fan of a systematic underestimation of kids.

But this isn't the question here, as I also mentioned that ##AB-BA= (AB-B)(AB-A)-(AB-A)(AB-B)=\ldots## can be used, too.

fresh_42 said:
When tutoring school kids, I frequently get to the problem of how to solve a linear equation system with two variables and two equations. They teach three procedures here, which are called: addition method, substitution method, and equality setting method. I find this stupid.
No doubt, since you have spent years studying mathematics, and you understand that using row operations on matrices is just a bookkeeping shortcut for what they're calling the addition method.

But almost certainly the kids you're tutoring haven't spent years and years in mathematics classes, and if they're just learning these simple methods of working with equations, they probably haven't been introduced to matrices yet. You didn't mention what grades these "school kids" were in, so I don't know if they are high school students or in the first year or two of college/university.

People have to crawl before they walk, and walk before they run. Teaching them matrix methods for solving a system of equations, without their having an understanding of what the matrix methods are doing isn't doing them any favors.

When you're helping people, you have to meet them where they are.

Mark44 said:
No doubt, since you have spent years studying mathematics, and you understand that using row operations on matrices is just a bookkeeping shortcut for what they're calling the addition method.

But almost certainly the kids you're tutoring haven't spent years and years in mathematics classes, and if they're just learning these simple methods of working with equations, they probably haven't been introduced to matrices yet. You didn't mention what grades these "school kids" were in, so I don't know if they are high school students or in the first year or two of college/university.

People have to crawl before they walk, and walk before they run. Teaching them matrix methods for solving a system of equations, without their having an understanding of what the matrix methods are doing isn't doing them any favors.

When you're helping people, you have to meet them where they are.
Sir can you suggest me any method for doing this problem using properties of matrices.For eg-by transpose or squaring or any other.Actually i am not taught about commutator method as i already mentioned earlier,so I cannot use this in my school.Please help.

StoneTemplePython said:
my take:
selecting ##A=0## and ##B=0## obviously satisfies the equation which knocks out choice (iii)

now consider a few basic things:
##AB = -(A+B)##
##(AB)^2 = (-1)^2(A+B)^2 = A^2 + AB +BA +B^2##
simplify for case (i) and case (ii)

if you go back to the original problem and square it you get
##\big(AB +A+B\big)^2 = \mathbf 0##
so
##(AB)^2 +A^2+B^2 +ABA + AB^2 +A^2B+ BAB + AB +BA= \mathbf 0##
##\big\{(AB)^2 +(A^2+B^2)\big\} +(ABA + A^2B)+ (AB^2 + BAB) + (AB +BA)= \mathbf 0##
simplify that.

in case (i) it should imply nilpotence of ##(AB)## and in case (ii) it should imply idempotence of ##(-AB)## -- which in combination with commuting matrices means simulataneously diagonalizable triangularizable (and nice trace properties with rank).

the intersection of (i) and (ii) then is when ##AB## is the zero matrix because otherwise nilpotent matrices are not diagonalizable.

I still don't have a particularly clean finish, but you might try solving this by assuming ##A## and ##B## are diagonal matrices and seeing if any complex numbers fit this bill. Scalar field was not stated nor were any relevant equations given so I'm guessing at what is useable, though ideas related to commuting and working in ##\mathbb C## seems reasonable to get going.

- - - -
edit:
if you work through case (ii) of commuting, carefully, what you get is a nice blocked structure, after similarity transform
where ##n=3 ##
##m \in\{0,1,..n\}##

 ##(AB) = \begin{bmatrix} -I_m & \mathbf 0 & \\ \mathbf 0& \mathbf {00}^T_{n-m} \end{bmatrix} = A \cdot B = \begin{bmatrix} J_m & \mathbf 0 & \\ \mathbf 0& N^{(A)}_{n-m} \end{bmatrix} \begin{bmatrix} -J_m^{-1} & \mathbf 0 & \\ \mathbf 0& N^{(B)}_{n-m} \end{bmatrix}##
with ##N^{(B)}_{n-m}, N^{(A)}_{n-m}## being nilpotent and commuting and obeying
##N^{(B)}_{n-m}N^{(A)}_{n-m} = \mathbf {00}^T_{n-m}##

and the matrices ##J_m## having non-zero eigenvalues.

This is reducible in blocked structure and if we ignore the nilpotent piece, we see that that ##B## is in effect the negative inverse of ##A## (sometimes called a pseudo inverse)

.. I of course agree with Fresh that this problem seems to be designed to look up the definition of a commutator and applying it, though OP seems quite against doing so -- I dropped in some other ideas accordingly.

also there's a link with Fibonnaci numbers in here...
Thanks i got it by the first method.

Physics lover said:
Thanks i got it by the first method.
(A+I)(B+I)=I
So A+I and B+I are inverses, so commute.

ehild, StoneTemplePython and fresh_42
Mark44 said:
When you're helping people, you have to meet them where they are.
High schoolers.

My point is that they are used to learn recipes. Easy steps to come to the solution. And

1. Bring the equations in the form ##ax+by=s\, , \,cx+dy=t##
3. If ##D\neq 0## then compute ##D^{-1}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}s\\t\end{bmatrix}##

is unbeatable in simplicity, error minimization, and to get to a solution without understanding the method. Of course I usually end up saying: Compute ##x=\frac{s}{a}-\frac{b}{a}y## and put it into the second equation. My problem with this 3 method teaching is, that a) the "correct" way isn't among them and b) they usually do not understand why they are choosing which method either, except that the exercise tells them which method to apply. So given this state of ignorance, they could as well mechanically build the inverse matrix. Sure, I try my best to explain why which method fits in which situation, but they are often only interested in "What shall I do?" and not in "Why am I doing this?". Kids with the patience for the latter question normally don't need tutoring.

haruspex said:
(A+I)(B+I)=I
So A+I and B+I are inverses, so commute.
Oh dear, that was the Gordian knot!

fresh_42 said:
My point is that they are used to learn recipes. Easy steps to come to the solution.
Surely you would agree that memorizing "cookbook" formulas is not the best way to actually learn a subject. My point, which I've already stated, is that merely manipulating symbols by doing row operations on a matrix without understanding how these row operations relate to the equations of a system, is a terrible way to learn.

Mark44 said:
Surely you would agree that memorizing "cookbook" formulas is not the best way to actually learn a subject. My point, which I've already stated, is that merely manipulating symbols by doing row operations on a matrix without understanding how these row operations relate to the equations of a system, is a terrible way to learn.
I do agree. I just doubt that it can always be achieved.

## 1. What is the condition for AB+A+B=0 where A and B are matrices?

The condition for AB+A+B=0 where A and B are matrices is that A and B must be square matrices of the same size, and the product AB must be equal to the negative of the sum of A and B.

## 2. Can AB+A+B=0 be true for non-square matrices?

No, the condition for AB+A+B=0 only holds for square matrices. For non-square matrices, the equation may have a solution, but it is not guaranteed to be equal to 0.

## 3. Is the condition for AB+A+B=0 satisfied if A and B are not of the same size?

No, the condition for AB+A+B=0 only holds if A and B are square matrices of the same size. If they are not of the same size, the equation is not defined.

## 4. Can the condition for AB+A+B=0 hold for non-commutative matrices?

Yes, the condition for AB+A+B=0 can hold for non-commutative matrices. As long as A and B are square matrices of the same size and the product AB is equal to the negative of the sum of A and B, the equation is satisfied.

## 5. What does the condition for AB+A+B=0 imply about the relationship between A and B?

The condition for AB+A+B=0 implies that A and B have some sort of inverse relationship. This can be seen by rearranging the equation to AB=-A-B, which can also be written as A(B+I)=-B. This suggests that A is the negative of the inverse of B, and vice versa.

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