A manipulative matrix question

[Raghav Gupta]

Homework Statement

If A and B are 2 matrices such that AB = A and BA =B, then B² is equal to
B
A
Zero matrix
I

Homework Equations

We can pre or post multiply a matrix on both sides of equation.

The Attempt at a Solution

(AB).(BA) = A.B
AB²A = A.B
Pre multiply both sides by A^-1
We get B²A = AB
Implies B²AB = AB as (A = AB)
Post multiplying by B^-1 and A we get B² = I

Is this correct?

 

[SammyS]

Homework Statement

If A and B are 2 matrices such that AB = A and BA =B, then B² is equal to
B
A
Zero matrix
I

Homework Equations

We can pre or post multiply a matrix on both sides of equation.

The Attempt at a Solution

(AB).(BA) = A.B
AB²A = A.B
Pre multiply both sides by A^-1
We get B²A = AB

Shouldn't that last equation be as follows?

B²A = B

By the way: Are you certain that matrix, A, has an inverse?

Implies B²AB = AB as (A = AB)
Post multiplying by B^-1 and A we get B² = I

Is this correct?

Are you certain that matrix, A, has an inverse?

 

[Mark44]

I don't see anything in the problem that says that either matrix is invertible, so there's no justification for assuming that A^-1 exists.

I found it helpful to start with B² and work with that.

 

[scientific601]

Maybe it was a trick question - nothing involving inverse matrices. Both A and B could be identity matrices. This would satisfy the problem statement in the strictest sense.

 

[SammyS]

I found it helpful to start with B² and work with that.

I started with A² .

 

[Mark44]

I started with A² .

That seems the long way around if the goal it to find B², but then I don't know what you did.

 

[SammyS]

That seems the long way around if the goal it to find B², but then I don't know what you did.

Well, I suppose we'll have to see what OP does before we can all compare notes.

 

[AdityaDev]

If you start with $B^2$,
Then i got A=I.
$B^2=B.B=B.BA=B^2A$
$B^2=B^2A$
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.

 

[SammyS]

If you start with $B^2$,
Then i got A=I.
$B^2=B.B=B.BA=B^2A$
$B^2=B^2A$
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.

While I satisfied that equation, it's not necessary that A = I .

 

[Mark44]

If you start with $B^2$,
Then i got A=I.
$B^2=B.B=B.BA=B^2A$
$B^2=B^2A$
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.

As Sammy said, B isn't necessarily I. Here's a counterexample, using your assumption that A = I.
$$B = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$
This matrix satisfies B² = B²A = B²I, yet $B \neq I$.

 

[Raghav Gupta]

Shouldn't that last equation be as follows?

B²A = B (Yes, sorry )

By the way: Are you certain that matrix, A, has an inverse? (No)

Are you certain that matrix, A, has an inverse? (No, as Mark44 10th post example proves that matrix B has determinant=0)

So,
I started with B² ,

$B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B$ Is multiplication of matrices associative which I am doing here?

Starting with A².
Following the above similar pattern , I get A²= A.
How did you Sammy get B² value by starting from A²?

 

[Dick]

So,
I started with B² ,

$B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B$ Is multiplication of matrices associative which I am doing here?

Starting with A².
Following the above similar pattern , I get A²= A.
How did you Sammy get B² value by starting from A²?

Yes, multiplication of matrices is certainly associative.

 

[Raghav Gupta]

Yes, multiplication of matrices is certainly associative.

How?

 

[Dick]

How?

If you don't want to take my word for it, https://proofwiki.org/wiki/Matrix_Multiplication_is_Associative

 

[Raghav Gupta]

If you don't want to take my word for it, https://proofwiki.org/wiki/Matrix_Multiplication_is_Associative

Oh, I was expecting a easy proof for it. I think we have to go in deep for that topic to understand the proof?

 

[Dick]

Oh, I was expecting a easy proof for it. I think we have to go in deep for that topic to understand the proof?

It's not deep, you just have to pay attention to the indices. Ignore the ring stuff. Just use that for real numbers (ab)c=a(bc).

 

[Raghav Gupta]

It's not deep, you just have to pay attention to the indices. Ignore the ring stuff. Just use that for real numbers (ab)c=a(bc).

Thanks, got it.

 

[SammyS]

So,
I started with B² ,

$B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B$ Is multiplication of matrices associative which I am doing here?

Starting with A².
Following the above similar pattern , I get A²= A.
How did you Sammy get B² value by starting from A²?

Of course, I didn't follow that pattern, but, yes, A²= A.

I start with:
A² = A²

BAAB = BAAB

((BA)A)B = (BA)(AB)

(BA)B = BA

BB = BAdded in Edit:
Here are two such matrices.
$\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]$

$\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]$

 

[Raghav Gupta]

I start with:
A² = A²

BAAB = BAAB

How you have written A² = BAAB ?

 

[SammyS]

Here are two such matrices.
$\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]$

$\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]$

How you have written A² = BAAB ?

I didn't say that.

I said

BAAB = BAABNow that I look at that, I only needed:

BAB = BAB

(BA)B = B(AB)

BB = BA

B² = B

 

[Raghav Gupta]

BAAB = BAABNow that I look at that, I only needed:

BAB = BAB

How you have written A² = A ?

 

[SammyS]

How you have written A² = A ?

I didn't use that at all. (I did mention in Post #18 that it is also true.)That post (#20) says, instead of starting with the identity BAAB = BAAB,

I could have started with BAB = BAB .

Either of these equations is obviously true.

You seem to not be following my method of working with these equations.

I manipulate each side of the equation independently.

I could have also done it this way.

BAB = (BA)B

= BB
=B²​

BAB = B(AB)

=BA
=B​

Therefore, B² = B .

Added In Edit:
I left the 2 out of the [ SUP][ /SUP]. DUH !

(Well the little icon has X², so why no 2 ?)

 

[Raghav Gupta]

I was simply stating that similar to B² = B, it's also true that A² = A.

I know that already and I have solved it as you can see in my earlier post.
What I am asking is you have told that earlier that you started with A² and when deriving that value of B² from A² you don't know that A² = A.
So how from A²
BAAB comes?

 

[Mark44]

I didn't use that at all. (I did mention in Post #18 that it is also true.)That post (#20) says, instead of starting with the identity BAAB = BAAB,

I could have started with BAB = BAB .

Either of these equations is obviously true.

You seem to not be following my method of working with these equations.

I manipulate each side of the equation independently.

I could have also done it this way.

BAB = (BA)B

= BB
=B²​

BAB = B(AB)

=BA
=B​

Therefore, B = B .

I'm pretty sure that you meant Therefore, B² = B.

 

[SammyS]

I know that already and I have solved it as you can see in my earlier post.
What I am asking is you have told that earlier that you started with A² and when deriving that value of B² from A² you don't know that A² = A.
So how from A²
BAAB comes?

I said I started with A², then I revised that (in the post) to say Instead start with BAAB. (Of course that's BA²B .)

After that, I further made the observation that I might just have well started with BAB .

I don't use A² = A at all.

 

[Raghav Gupta]

Okay, got it.
Thanks to all of you.
Learnt quite a lesson from this question.

 

[Ray Vickson]

Okay, got it.
Thanks to all of you.
Learnt quite a lesson from this question.

The easiest solution of all is to start with AB = A and BA =B, and multiply both sides of the first equation on the left by B, to get B(AB) = BA, or (BA)B = B = B*B because BA=B. That seems to me to be about the simplest solution possible.

 

[Raghav Gupta]

Thanks to you as well Ray.