# matrix question: AB = B

• Jul 8th 2009, 03:47 PM
chrisc
matrix question: AB = B
This is an exam question I had. I think I deserve full marks, but the prof is hard to speak too because of his english and I cant seem to find a way to convince him why I deserve full marks. So I want a second opinion, before I go to the department head.

Let A and B be nxn matrices. Can we deduce from AB=B that B = 0 or A = In?

Let B = 0.
AB = B
A0 = 0
0 = 0
therefore we can deduce that AB=B when B = 0

Let A = In
AB = B
[1,0 - 0,1]B = B
B = B
thefefore we can deduce that AB=B when A = In

(that junk in the squar bracket is an identity matrix)
(for each question, I showed each matrix as a general 2x2 matrix and expanded them to help get my message across)

From what I can understand what his complain was, is that I worked the problem out backwards. If someone can clarify this for me, give me an opinion, please let me know.
I only got 3/6 for the question. However, the exam is only out of 37 so each mark is worth quite a bit.
• Jul 8th 2009, 04:25 PM
halbard
Have a look at the equation

$\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix }0&0\\0&1\end{pmatrix}
=\begin{pmatrix}0&0\\0&1\end{pmatrix}$
.

This has the form $AB=B$. Does that answer your question?

Do you need to deduce from $AB=B$ that $A=I$ or that $B=O$? Or do you need to deduce from $A=I$ or $B=O$ that $AB=B$?

Which is it?
• Jul 9th 2009, 01:28 PM
TheAbstractionist
Any idempotent 2×2 matrix $A$ (i.e. a matrix $A$ such that $A^2=A)$ that is not the zero or the identity matrix will serve as a counterexample setting $B=A.$

Example: $A=B=\begin{pmatrix}1&0\\0&0\end{pmatrix}.$ There are certainly plenty of nonzero nonidentity 2×2 idempotent matrices out there.
• Jul 9th 2009, 01:31 PM
chrisc
After talking with him more, this is what I found out he wanted.

AB=B
AB - B = 0
B(A-In) = 0
therefore either B = 0 or A = In

hehe, so simple
• Jul 9th 2009, 01:52 PM
Plato
Quote:

Originally Posted by chrisc
After talking with him more, this is what I found out he wanted.
AB=B
AB - B = 0
B(A-In) = 0
therefore either B = 0 or A = In

It may well be what was wanted. But it is completely wrong.
Consider this counter example.
P.S.
Quote:

Originally Posted by chrisc
This is an exam question I had. I think I deserve full marks, but the prof is hard to speak too because of his english and I cant seem to find a way to convince him why I deserve full marks. So I want a second opinion, before I go to the department head.

Given that counter-example, I would certainly go to the department chair with such a complaint.
If you are sure that you have understood him correctly, that is ‘Is the quote exactly what you were told?”
As a retired department chair, I can tell you that this sort of concern needs to be reported.
But that said, you need to be quite sure that you have all the facts straight.
Did you quote the instructor exactly and fairly?
• Jul 9th 2009, 02:44 PM
CaptainBlack
Quote:

Originally Posted by chrisc
After talking with him more, this is what I found out he wanted.

AB=B
AB - B = 0
B(A-In) = 0
therefore either B = 0 or A = In

hehe, so simple

Have you missed something out of the question per chance?

CB
• Jul 10th 2009, 07:46 AM
chrisc
Quote:

Originally Posted by Plato
Did you quote the instructor exactly and fairly?

He was speaking broken English, and could barely understand me. What should of been a 1 minute conversation turned out to be 10.
I will try to ask him for the exact answer, and get something directly out of him. But up to this point, I am quite sure that he said that the answer I posted above is what he wanted.
• Jul 15th 2009, 12:49 AM
Gamma
I think the question has been fully answered by everyone, but I was just thinking it might be some helpful advice for you in future exams. If you are trying to prove that either A or B is true, you cannot just do what you did. In english this word "or" is kind of ambiguous, but in mathematics it has a very precise meaning.

Either A is true, or B is true, or both is true. You cannot just check that both A and B work, it is much more restrictive than this. The statement is not true if for instance some other option C was true. To prove such a thing you need to show one of two things: (1) if A is not true, then B is true (2)if B is not true, then A is true.

For instance if I say vanilla and chocolate are ice cream choices, can you deduce that if someone has ice cream, then they have chocolate or vanilla? sure, if they have chocolate ice cream they have ice cream. If they have vanilla ice cream they have ice cream. But someone could also have strawberry ice cream and still have ice cream, so you cannot deduce that statement is true.

My point is even though its clear the statement is not true, even if it were true you have not proven it at all.

Further pondering:
That proof is legit in an integral domain, correct? well except the whole lack of regard for noncommutativity of the matrices?

Like it is true that if there are no zero divisors $ab=b \Leftrightarrow ab-b=0 \Leftrightarrow (a-1)b=0 \Leftrightarrow a=1$ or $b=0$

Which I guess explains why the only idempotent elements of an integral domain are the multiplicative and additive identities.

That blows my mind that the prof thought this statement was actually true, he should know better. Also that no one in the class came up with a counter-example to make him realize his mistake.
• Oct 10th 2009, 07:03 AM
HallsofIvy
Quote:

Originally Posted by chrisc
He was speaking broken English, and could barely understand me. What should of been a 1 minute conversation turned out to be 10.
I will try to ask him for the exact answer, and get something directly out of him. But up to this point, I am quite sure that he said that the answer I posted above is what he wanted.

Do YOU understand that, whether it was what he wanted or not, that is completely wrong? Given that AB= B, you CANNOT conclude that B= 0 of A= I. You have now been shown many counter examples.

By the way, the proof you posted originally, that you felt "deserve full marks" was the wrong way. You proved "If A= I or B= 0 then AB= B". You claimed you were proving "If AB= B then A= I or B= 0"- which, once again, is NOT true.

As far as "AB= B so AB- B= (A- I)B= 0" is concerned, it does NOT follow that either A- I= 0 or B= 0. Matrix multiplication has "zero divisors"- that is, there exist NON-ZERO matrices A and B such that AB= 0. "If AB= 0 then A= 0 or B= 0" is NOT true for matrices.
• Oct 10th 2009, 08:08 AM
Swlabr
Quote:

Originally Posted by Gamma
That blows my mind that the prof thought this statement was actually true, he should know better. Also that no one in the class came up with a counter-example to make him realize his mistake.

I would agree - I suspect we are missing something from the question. As I understand it, exams are checked over by a secondary person to stop these kind of problems (but then again, I tend to think everything is perfect).

Is the result not true if we fix $A$ and it holds for all $B$ (as in, if $AB=B$ for all $B \in R$ then $A =1$), and similarly if $AB=B$ for all $A \in R$ then $B=0$ (by uniqueness of identities and zeros). So could the way the question was asked perhaps not be a variation on this?
• Oct 11th 2009, 08:49 AM
HallsofIvy
Quote:

Originally Posted by Swlabr
I would agree - I suspect we are missing something from the question. As I understand it, exams are checked over by a secondary person to stop these kind of problems (but then again, I tend to think everything is perfect).

Is the result not true if we fix $A$ and it holds for all $B$ (as in, if $AB=B$ for all $B \in R$ then $A =1$), and similarly if $AB=B$ for all $A \in R$ then $B=0$ (by uniqueness of identities and zeros). So could the way the question was asked perhaps not be a variation on this?

If B is invertible then AB= B gives A= I by multiplying on the right by $B^{-1}$ so, yes, if AB= B for all B (and so some B is invertible) then A= I.

If AB= B for all A, then AB- A= A(B- I)= 0. Now, there exist invertible A so that B-I= 0 and B= I.

I suspect that chrisc has simply made a hash of the problem.
• Oct 11th 2009, 09:07 AM
Swlabr
Quote:

Originally Posted by HallsofIvy
If AB= B for all A, then AB- A= A(B- I)= 0. Now, there exist invertible A so that B-I= 0 and B= I.

Your " $AB- A= A(B- I)= 0$" should be $AB-B=(A-I)B=0$...

The fact that it only holds for $B=0$ can be seen by showing that it holds for zero, then setting $A=0$: $B=AB=0B=0$.
• Oct 11th 2009, 10:03 AM
tonio
Quote:

Originally Posted by chrisc
This is an exam question I had. I think I deserve full marks, but the prof is hard to speak too because of his english and I cant seem to find a way to convince him why I deserve full marks. So I want a second opinion, before I go to the department head.

Let A and B be nxn matrices. Can we deduce from AB=B that B = 0 or A = In?

Let B = 0.
AB = B
A0 = 0
0 = 0
therefore we can deduce that AB=B when B = 0

Let A = In
AB = B
[1,0 - 0,1]B = B
B = B
thefefore we can deduce that AB=B when A = In

(that junk in the squar bracket is an identity matrix)
(for each question, I showed each matrix as a general 2x2 matrix and expanded them to help get my message across)

From what I can understand what his complain was, is that I worked the problem out backwards. If someone can clarify this for me, give me an opinion, please let me know.
I only got 3/6 for the question. However, the exam is only out of 37 so each mark is worth quite a bit.

In my opinion, and without intending the slightest ofense, you not only don't deserve full marks but not mark at all: you were supposed to find out whether AB = B ==> B = 0 ir A = I_n, and you gave examples when A, B are of this or that form then AB = B, and this is NOT what you were asked to!

In fact, the answer should be simply NO, since we can have AB = B
without B being the zero matrix or A the identity one.

It is my opinion that if you got a mark 3/6 you better count your blessings and shut the hell up, lest some other lecturer re-checking your exam will lower your mark, as I undoubtedly would.

Tonio