Is it ok to assume matrices A and B as identity matrix?

[vcsharp2003]

Homework Statement

4. If $A$ and $B$ are two matrices such that $AB = B$ and $BA = A$, then $A^2 + B^2$ equals:

(a) $A + B$
(b) $2BA$
(c) $2AB$
(d)$BA$

Relevant Equations

$AI = IA = A$ where $I$ is identity matrix and $A$ is any square matrix whose product with identity matrix is defined

Since $AB = B$, so matrix $A$ is an identity matrix.
Similarly, since $BA = A$ so matrix $B$ is an identity matrix.

Also, we can say that $A^2 = AA=IA= A$ and $B^2 = BB=IB= B$.
Therefore, $A^2 + B^2 = A + B$ which means (a) is a correct answer.

Also we can say that $A^2 + B^2 = I^2 + I^2 = II + II = AB + AB = 2AB$,
and that $A^2 + B^2 = I^2 + I^2 = II + II = BA + BA = 2BA$. From these conclusions, it also follows that (b) and (c) are correct answers.

Thus, according to me (a),(b) and (c) are correct answers. But the correct answer is given as (a) only.

 

[DrClaude]

Since $AB = B$, so matrix $A$ is an identity matrix.
Similarly, since $BA = A$ so matrix $B$ is an identity matrix.

This is not true. Counter example:
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad \quad B = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
$$

 

[FactChecker]

This is not true. Counter example:
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad \quad B = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
$$

Good example. A trivial example is A=B=0. Your example is much more interesting and informative.

 

[vcsharp2003]

This is not true. Counter example:
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad \quad B = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
$$

I see. So, how would I go about doing this question since A or B are not necessarily identity matrix.

 

[DrClaude]

To complement my answer: it is important to parse carefully problem statements.

Homework Statement: 4. If A and B are two matrices such that $AB = B$ and $BA = A$, then $A^2 + B^2$ equals:

(a) $A + B$
(b) $2BA$
(c) $2AB$
(d)$BA$

Here, you have two given matrices. $A$ would be the identity if $AB = B$ were true for any matrix $B$.

 

[DrClaude]

I see. So, how would I go about doing this question since A or B are not necessarily identity matrix.

Expand the squares and make some substitutions using the given equalities.

 

[fresh_42]

You can conclude the right solution without calculating anything.

To prove the right answer, start with $A^2=A\cdot A.$

 

[vcsharp2003]

You can conclude the right solution without calculating anything.

To prove the right answer, start with $A^2=A\cdot A.$

$A^2 = A \cdot A = AB \cdot A = A \cdot (BA) = AB$ and $B^2 = B \cdot B = BA \cdot B = B \cdot (AB) = BA$.

Sorry, I couldn't get it using above steps, since I'm back to where I started from.

 

[vcsharp2003]

Expand the squares and make some substitutions using the given equalities.

I have tried that a few times but somehow still don't get any of the answers. I'll keep trying.

 

[vcsharp2003]

To complement my answer: it is important to parse carefully problem statements.Here, you have two given matrices. $A$ would be the identity if $AB = B$ were true for any matrix $B$.

I see. This is a tough question.

 

[DrClaude]

$A^2 = A \cdot A = AB \cdot A = A \cdot (BA) = AB$ and $B^2 = B \cdot B = BA \cdot B = B \cdot (AB) = BA$.

Sorry, I couldn't get it using above steps, since I'm back to where I started from.

So you have
$$
A^2 + B^2 = AB + BA
$$
Use the equalities one more time.

 

[fresh_42]

$A^2 = A \cdot A = AB \cdot A = A \cdot (BA) = AB$ and $B^2 = B \cdot B = BA \cdot B = B \cdot (AB) = BA$.

Sorry, I couldn't get it using above steps, since I'm back to where I started from.

$A^2=A\cdot A= A\cdot (BA)= (AB)\cdot A= \ldots$

 

[vcsharp2003]

So you have
$$
A^2 + B^2 = AB + BA
$$
Use the equalities one more time.

Ok. So, substituting in RHS from the given equations, we get
$$A^2 + B^2 = AB + BA = B + A$$
$$ \therefore \text {(a) is the answer} $$

 

[fresh_42]

Ok. So, substituting in RHS from the given equations, we get
$$
A^2 + B^2 = AB + BA = B + A
$$

$$ \therefore \text {(a) is the answer} $$

Yes. And what is the correct argument without calculation?

 

[vcsharp2003]

Yes. And what is the correct argument without calculation?

Sorry, I didn't get your question.

 

[fresh_42]

Sorry, I didn't get your question.

You can "see" directly from the problem statement that (a) is the correct answer without multiplying anything; just by inspection.

 

[vcsharp2003]

You can "see" directly from the problem statement that (a) is the correct answer without multiplying anything; just by inspection.

I cannot see it.
Also, how can we be sure that other options are not correct?

 

[fresh_42]

I cannot see it.
Also, how can we be sure that other options are not correct?

Well, it is easy to see that any two cannot be simultaneously correct since they are all different (unless $A= 0$ or $B=0.$ I assumed that neither is the zero matrix.
a) $A+B$
b) $2BA = 2 A$
c) $2AB =2 B$
d) $BA=A$

However, there is one property that distinguishes a) from all the rest! It is important to learn to see it. So take a few minutes to think about it.

 

[fresh_42]

HINT The fact that it shows a to be distinguished does not guarantee the answer. But you can trivially consruct counterexamples for b,c,d once you see it.

That's new to me. I'm looking forward to the answer. My solution is a guarantee that a) is the only possible solution besides $A=B=0$ when all 4 answers are true.

 

[vcsharp2003]

Well, it is easy to see that any two cannot be simultaneously correct since they are all different (unless $A= 0$ or $B=0.$ I assumed that neither is the zero matrix.
a) $A+B$
b) $2BA = 2 A$
c) $2AB =2 B$
d) $BA=A$

However, there is one property that distinguishes a) from all the rest! It is important to learn to see it. So take a few minutes to think about it.

I get the first part where clearly b,c and d cannot be correct. But, the second part is too tricky for me and the only thing that comes to mind is that A and B are of the same order in a, whereas, in b,c,d A and B need not be of the same order as they involve multiplication.

 

[hutchphd]

Yeah I was wrong. So it goes. I need a quicker delete finger!

 

[vcsharp2003]

That's new to me. I'm looking forward to the answer. My solution is a guarantee that a) is the only possible solution besides $A=B=0$ when all 4 answers are true.

If A and B are identity matrices as per my OP, then also they i.e. a,b,c are true, but not d.

 

[fresh_42]

I get the first part where clearly b,c and d cannot be correct. But, the second part is too tricky for me and the only thing that comes to mind is that A and B are of the same order in a, whereas, in b,c,d A and B need not be of the same order as they involve multiplication.

Yes, that is true. At least in principle. I would have phrased it differently:

The problem statement is perfectly symmetric in $A$ and $B$. Means: We can exchange the roles of $A$ and $B$ and get the exact same problem. But only the answer a) is also symmetric in $A$ and $B$, hence all others are impossible.

d) is not a possible solution for the identity matrix, but yes, you are right, $A=B=I$ yields a),b),c) as correct solutions.

 

[WWGD]

I guess , to be somewhat pedantic, it's more of an issue about stabilizer of A,B.

https://mathworld.wolfram.com/Stabilizer.html

 

[WWGD]

notice we can't have A=B nor AB=BA, otherwise options b,c would be identical. That discounts either being the identity as well.

 

[martinbn]

Square $A=BA$, to get $A^2=BABA=B(AB)A=B(B)A=B(BA)=BA=A$. Same way $B^2=B$. Then $A^2+B^2=A+B$.

Of course what @fresh_42 said: the expression is symmetric for $A$ and $B$, and only one answer is.