Difficult Matrix Identity Question

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Homework Help Overview

The discussion revolves around evaluating the truth of several matrix identities involving n × n matrices A, B, and C. Participants are tasked with determining whether specific statements about matrix operations hold true universally for all matrices of the given size.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of distributive rules to evaluate both sides of the matrix equations. There are attempts to analyze specific statements, particularly (a) and (b), by expanding expressions and checking for equality. Some participants express uncertainty about the validity of their evaluations and seek clarification on how to rigorously prove their findings.

Discussion Status

Some participants have identified potential falsehoods in statements (a) and (b) but have not reached a definitive conclusion. There is ongoing exploration of specific examples and the application of matrix multiplication properties. Guidance has been offered regarding the evaluation process, but no consensus has been reached on the truth of the statements.

Contextual Notes

Participants are reminded that the statements must be evaluated for all n × n matrices, not just specific cases. There is an emphasis on the importance of rigor in mathematical proofs and the need to consider the properties of matrix multiplication.

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Homework Statement



For each of the following statements, select whether the statement is true or false for all n × n matrices A, B , C. (Note that you are being asked whether the statement is true or false for all n × n matrices A, B, C, not just for some A, B, C.)

a) (-6 A - 4 B)2 = 48 AB + 36 A2 + 16 B2


b) (6 I + A2)(-4 I + A2) = (-4 I + A2)(6 I + A2)

c) A = ±8 B ⇒ A2 = 64 B2

d) (A(-4 I - 6 A))T = AT(-4 I - 6 A)T

The Attempt at a Solution



What I know is (b) and (d) are false.

(b) is false because AB is not equal to BA unless the matrix is invertible.

(d) is false because (AB)T = BTAT.
 
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In a and b, it's a good idea to use the distributive rules ( A(B+C)=AB+AC and (A+B)C=AC+BC ) to evaluate both the left-hand side and the right-hand side. Have you tried this?

Good thinking in d, but this is only a reason to suspect that d is false. It doesn't prove it conclusively. What do you get if you evaluate the right-hand side using the distributive rule and that ##(A+B)^T=A^T+B^T##? Is the answer different if you first change the order of the two factors on the right-hand side?
 
Fredrik said:
In a and b, it's a good idea to use the distributive rules ( A(B+C)=AB+AC and (A+B)C=AC+BC ) to evaluate both the left-hand side and the right-hand side. Have you tried this?

Good thinking in d, but this is only a reason to suspect that d is false. It doesn't prove it conclusively. What do you get if you evaluate the right-hand side using the distributive rule and that ##(A+B)^T=A^T+B^T##? Is the answer different if you first change the order of the two factors on the right-hand side?

No I havent. How should I go about evaluating (a) and (b) using the principle you've provided.

(a): (-6 A - 4 B)2 = 48 AB + 36 A2 + 16 B2

(-6 A - 4 B)(-6A -4B) =? 48AB + 36AA + 16BB

How should I take it further?
 
negation said:
No I havent. How should I go about evaluating (a) and (b) using the principle you've provided.

(a): (-6 A - 4 B)2 = 48 AB + 36 A2 + 16 B2

(-6 A - 4 B)(-6A -4B) =? 48AB + 36AA + 16BB

How should I take it further?
You don't. This is where you stop, because what you just did proves that a is true. (Edit: My bad. See the correction below). You could however have included more information (all the steps) in your calculation:
$$(-6A-4B)^2= (-6A-4B)(-6A-4B) = (-6A)(-6A-4B)+(-4B)(-6A-4B)=\cdots$$ You should do problem b this way. You don't have to tell me all the steps, but you should evaluate both the left-hand side and the right-hand side this way, and post the results. (If the result is wrong, I may ask you to post the calculation).
 
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Fredrik said:
You don't. This is where you stop, because what you just did proves that a is true. You could however have included more information (all the steps) in your calculation:
$$(-6A-4B)^2= (-6A-4B)(-6A-4B) = (-6A)(-6A-4B)+(-4B)(-6A-4B)=\cdots$$ You should do problem b this way. You don't have to tell me all the steps, but you should evaluate both the left-hand side and the right-hand side this way, and post the results. (If the result is wrong, I may ask you to post the calculation).

Alright. Awesome. So (a) is true. You said I could have included further information (a more rigourous manner), I wish to know how.

(b): (6 I + A2)(-4 I + A2) = (-4 I + A2)(6 I + A2)

(6 I + AA)(-4I + AA) = (-4I + AA)(6I + AA)

but..

(24II + 6AAI -4IAA + AAAA) = 24I + 6AA - 4AA + A4


AB =/= BA

so (b) is false


Edit: I just saw.

(-6A-4B)^2= (-6A-4B)(-6A-4B) = (-6A)(-6A-4B)+(-4B)(-6A-4B) = 36AA + 24AB + 24AB +16BB = 36AA + 48AB +16BB
 
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Fredrik said:
You don't. This is where you stop, because what you just did proves that a is true. You could however have included more information (all the steps) in your calculation:
$$(-6A-4B)^2= (-6A-4B)(-6A-4B) = (-6A)(-6A-4B)+(-4B)(-6A-4B)=\cdots$$ You should do problem b this way. You don't have to tell me all the steps, but you should evaluate both the left-hand side and the right-hand side this way, and post the results. (If the result is wrong, I may ask you to post the calculation).

I think it's good.
 
negation said:
So (a) is true.
Oops. It's not. I didn't think it through. Sorry about that. 24AB+24BA (which appears in the left-hand side when you rewrite it as a sum of four terms) is not always equal to 48AB (which appears in the right-hand side). So we can conclude that (a) is very likely not true for all A,B. To complete the proof, you should look for a specific choice of A and B that makes the equality in (a) false.

negation said:
You said I could have included further information (a more rigourous manner), I wish to know how.
Your book should include a definition of matrix multiplication, a definition of the transpose operation, and several theorems that state elementary results like ##(A+B)C=AC+BC## and ##(A^T)^T=A##. One way to include more information is to use only one of those theorems at a time. But if you want to do things like (A+B)(C-D)=AC-AD+BC-BD in one step, that's OK too. You just need to make sure that your readers can tell what you're doing.

negation said:
(b): (6 I + A2)(-4 I + A2) = (-4 I + A2)(6 I + A2)

(6 I + AA)(-4I + AA) = (-4I + AA)(6I + AA)

but..

(24II + 6AAI -4IAA + AAAA) = 24I + 6AA - 4AA + A4AB =/= BA

so (b) is false
I'm not sure sure what you're doing here. If you just take the left-hand side ##(6I+A^2)(-4I+A^2)## and rewrite it as a sum of four terms, what do you get? Did you mean that you got ##24I+6A^2-4A^2+A^4##? In that case, there's a sign error. You can of course also simplify the two terms in the middle to ##2A^2##.

Do you get a different result for the right-hand side?
 
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Fredrik said:
Oops. It's not. I didn't think it through. Sorry about that. 24AB+24BA (which appears in the left-hand side when you rewrite it as a sum of four terms) is not always equal to 48AB (which appears in the right-hand side). So we can conclude that (a) is very likely not true for all A,B. To complete the proof, you should look for a specific choice of A and B that makes the equality in (a) false.Your book should include a definition of matrix multiplication, a definition of the transpose operation, and several theorems that state elementary results like ##(A+B)C=AC+BC## and ##(A^T)^T=A##. One way to include more information is to use only one of those theorems at a time. But if you want to do things like (A+B)(C-D)=AC-AD+BC-BD in one step, that's OK too. You just need to make sure that your readers can tell what you're doing.I'm not sure sure what you're doing here. If you just take the left-hand side ##(6I+A^2)(-4I+A^2)## and rewrite it as a sum of four terms, what do you get? Did you mean that you got ##24I+6A^2-4A^2+A^4##? In that case, there's a sign error. You can of course also simplify the two terms in the middle to ##2A^2##.

Do you get a different result for the right-hand side?

Yes that's was what I meant. what do you mean by sign error?

(a) must be false. The system states so. But I do not know why it's false.

Let me rework it.

Edit: Ah I missed it. In ref to part(a), 24AB + 24 BA is not necessarily 48 AB as you've state. It could be but it is not a corollary that AB + BA = 2AB.
 
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Fredrik said:
Oops. It's not. I didn't think it through. Sorry about that. 24AB+24BA (which appears in the left-hand side when you rewrite it as a sum of four terms) is not always equal to 48AB (which appears in the right-hand side). So we can conclude that (a) is very likely not true for all A,B. To complete the proof, you should look for a specific choice of A and B that makes the equality in (a) false.


Your book should include a definition of matrix multiplication, a definition of the transpose operation, and several theorems that state elementary results like ##(A+B)C=AC+BC## and ##(A^T)^T=A##. One way to include more information is to use only one of those theorems at a time. But if you want to do things like (A+B)(C-D)=AC-AD+BC-BD in one step, that's OK too. You just need to make sure that your readers can tell what you're doing.


I'm not sure sure what you're doing here. If you just take the left-hand side ##(6I+A^2)(-4I+A^2)## and rewrite it as a sum of four terms, what do you get? Did you mean that you got ##24I+6A^2-4A^2+A^4##? In that case, there's a sign error. You can of course also simplify the two terms in the middle to ##2A^2##.

Do you get a different result for the right-hand side?

for part(b)

LHS = RHS
 
  • #10
Part(a) = false, part(b) = true, part(c) = true and verified true by the system.

(A(-4 I - 6 A))T = AT(-4 I - 6 A)T is true, but why?
 
  • #11
You know that ##(A(-4I-6A))^T)=(-4I-6A)^TA^T##. If you rewrite this right-hand side as a sum of two terms, and compare the result to what you get when you rewrite the right-hand side of the equality in your post as the sum of two terms, you will find that they're the same. So ##A^T## commutes with ##(-4I-6A)^T##. This is no surprise since ##A^T## commutes with ##A^T##, and ##I^T## commutes with everything.
 
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  • #12
fredrik said:
you know that ##(a(-4i-6a))^t)=(-4i-6a)^ta^t##. If you rewrite this right-hand side as a sum of two terms, and compare the result to what you get when you rewrite the right-hand side of the equality in your post as the sum of two terms, you will find that they're the same. So ##a^t## commutes with ##(-4i-6a)^t##. This is no surprise since ##a^t## commutes with ##a^t##, and ##i^t## commutes with everything.

(a(-4i-6a))t = lhs = (-4i -6a)tat
∴ lhs = (-4i -6a)tat

rhs = -4atit - 6atat = (-4it-6at)at = (-4i - 6a)tat
 
  • #13
That's good.

I missed your question about what sign error I was referring to earlier. I meant that when you evaluate ##(6I+A^2)(-4I+A^2)##, one of the terms is ##-24I##, not ##+24I##.
 
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