If A and B Are non-singular then we have determinant of (A^2-B^2)=-determinant of AB , Am I correct? If so, then the problem is If A and B are singular, how can we prove?
The relations given above clearly say that det ( A) = det (B). I do not know how you infered det(A^2-B^2) = - det(A. B) ? Here either both are singular or both are non-singular.
In the case of non-singular B the answer is det(A^2-B^2)=0.
In fact, multiplying A^2B=AB^2 from the left by A one gets A^3B=A^2B^2,
so B^4=A^2B^2. This implies (B^2-A^2)B^2 = 0, so either det(A^2-B^2)=0 or det(B)=0.
@ Mittal, I was thinking is it true that det(A^2-B^2) = - det(A. B)
Given your assumptions, the case that either B or A is non-singular can not happen.
Starting from A^3=B^3 multiply by B from the right and insert A^2B=AB^2. This gives A^2B^2=B^4. If B would be non-singular, we can obtain A^2=B^2, which is excluded in the assumption. By symmetry, the same statement is valid for A.
@ Sascha Kurz, I think it will be difficult now to find the determinant?.
So, the answer is det(A^2-B^2)=0. In fact, assume contrary that A^2-B^2 is invertible,
then the identity (B^2-A^2)B^2 = 0 implies B^2 = 0. The same way, multiplying A^2B=AB^2 from the right by B one gets A^2B^2 = AB^3 = A^4, so
A^2(A^2 - B^2) = 0, and our assumption leads to A^2 = 0. So, A^2 = B^2, which contradicts the initial assumptions of the problem.
That's it. Maybe this can serve as a nice brain-teaser in linear algebra courses.
So indeed one can write det(A^2-B^2) = - det(AB) as a more complicated and less meaningful way for just det(A^2-B^2)=0; given the stated assumptions.
Clearly det(A) =det(B)=0 since if either is invertible then the other must be as well (since A^3=B^3 so det(A)^3=det(B)^3) and so A^2B=AB^2 reduces to A=B which is a contradiction. Now, with det(A)=det(B)=0 and the below statement:
A^2(B^2-A^2)=0
{proof: A^2(B^2-A^2)=(A^2B)B-A(A^3)=AB^3-AB^3=0}
We have A^2B^2=A^4.
Therefore (A^2-B^2)^2=A^4-A^2B^2-B^2A^2+B^4=B^4-B^2A^2=B^2(B^2-A^2)
Finally [det(A^2-B^2)]^2=det[(A^2-B^2)^2]=det(B^2(B^2-A^2))=(detB)^2det(B^2-A^2)=0
because det(B)=0 so det(A^2-B^2)=0.
If A or B are nonsingular, then det (A^2 - B^2) = 0, since can be easily proved that
A^2 B^2 = A^4, or B^4. If A or B is singular, and since detA = det B, then, once again,
det (A^2 - B^2) = 0
From given assumption we can easily prove that A and B are sigular. After that we will take two example-:
1) first row and second row of A are (1 0) and (0 0) and first row and second row of B are (-1 0) and (0 0). In this case det ( A^2 - B^2) =0.
2) first row and second row of A are (1 0) and (0 0) and first row and second row of B are (0 0) and (0 1). In this case det ( A^2 - B^2) = -1.
Hence we can not say anything about the det(A^2 - B^2).
Dear Prasanth
Let's start with A2B=AB2. By rules of matrix computations we
get : Det(A)Det(A)Det(B)=Det(A)Det(B)Det(B), moreover
det(A)^3=det(B)^3 so either both are of full rank or neither is.
Then consider equation B(A2 - B2)A=BA2 -A2B=0 (BY ASSUMPTION BA^2=A^2B)
Again by basic rules of matrix comp.
Det(A)Det(A2 - B2)Det(B)=Det(A)2Det(A2 - B2)=0 (Since Det(A)=Det(B)),
O.k so after some thought it is quite clear from assumtions that A and B can't be of full rank so that det(A)=det(B)=0.
This is because the fact that since if det(A)=det(B) \neq 0 then this would imply the existence of the inverse of A and B so that the equation
A2 B=AB2 would result after multiplication by A-1 from the right and after multiplication by B-1 from the left to equation A=B, which would be a contradiction to the original claim A does not equal B. So the only possibility is det(A)=det(B)=0 ???
So looked like I only complicated things...
So in fact I think the last commentor Nitin Bisth is correct in his claim.
Nothing can be said about det(A^2-B^2).
Prasanth are you sure you have the facts dorrect ? Could you double chek them ? Of course it might be that my mind is again playing tricks on me :-).
best regards !
Samuli
Dear Dr. Prasanth G. Narasimha-Shenoi
Solution: Since
A³-B³-AB²+A²B=0
⇒ (A³-AB²)+(A²B-B³)=0
⇒ A(A²-B²)+(A²-B²)B=0
and
A³-B³-AB²+AB²=0
⇒ (A³-AB²)+(AB²-B³)=0
⇒ A(A²-B²)+(A-B)B²=0
We have
A(A²-B²)+(A²-B²)B=0=A(A²-B²)+(A-B)B²
⇒ (A²-B²)B=(A-B)B².
Therefore
det(A²-B²)det B = det(A-B)det B²
det(A²-B²) = (1/(detB))det(A-B)det B², if det B≠0
Hence
det(A²-B²) = det(A-B)det B, if det B≠0
@ Samuli Yeah the facts are correct. Sorry that I am not able to answer
- Badges
- Science topic
- Similar topics
- Journal Impact Factor