Proof Involving Matrix Polynomials and Matrix Multiplication

In summary: RGVIn summary, the property needed to prove the equation (A^n)B = B(C^n) is the associative property of matrix multiplication. By using induction and assuming the equation is true for n=k, we can prove that it is also true for n=k+1. Therefore, (A^n)B = B(C^n) is true for all n∈ℕ.
  • #1
Reckoning of Sand
13
0

Homework Statement



Let [itex]A[/itex] be an [itex]nxn[/itex] matrix, and [itex]C[/itex] be an [itex]mxm[/itex] matrix, and suppose [itex]AB = BC[/itex].

(a) Prove the following by induction: For every [itex]n∈ℕ[/itex], [itex](A^n)B = B(C^n)[/itex]. What property of matrix multiplication do you need to prove this?

Homework Equations



The four basic properties of matrix multiplication discussed in my course are

1. Distributive: [itex](A + B)C = AC + BC[/itex] and [itex]C(A + B) = CA + CB[/itex]
2. Scalar Commutativity: [itex](tA)B = t(AB) = A(tB)[/itex]
3. Associative: [itex]A(BC) = (AB)C[/itex]

The Attempt at a Solution



If [itex]n = 1[/itex], then [itex](A^n)B = B(C^n)[/itex] becomes [itex](A^1)B = B(C^1)[/itex] which is[itex] AB = BC[/itex] which is supposed to be true.

Assume [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k[/itex]. Then [itex](A^k)B = B(C^k)[/itex] is true.

Prove [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k + 1[/itex].

[itex](A[/itex]k+1[itex])B = B(C[/itex]k+1[itex]) ⇒ (A(A^k))B = B(C(C^k))[/itex]

I'm not really sure how to proceed. I tried rewriting the matrix products as the summations of the products of their entries, but that didn't seem to get me anywhere. I know that I have to relate the equation when [itex]n = k + 1[/itex] to the equation when [itex]n = k[/itex] as part of induction, but I'm still stuck.
 
Physics news on Phys.org
  • #2
Naridax said:

Homework Statement



Let [itex]A[/itex] be an [itex]nxn[/itex] matrix, and [itex]C[/itex] be an [itex]mxm[/itex] matrix, and suppose [itex]AB = BC[/itex].

(a) Prove the following by induction: For every [itex]n∈ℕ[/itex], [itex](A^n)B = B(C^n)[/itex]. What property of matrix multiplication do you need to prove this?

Homework Equations



The four basic properties of matrix multiplication discussed in my course are

1. Distributive: [itex](A + B)C = AC + BC[/itex] and [itex]C(A + B) = CA + CB[/itex]
2. Scalar Commutativity: [itex](tA)B = t(AB) = A(tB)[/itex]
3. Associative: [itex]A(BC) = (AB)C[/itex]

The Attempt at a Solution



If [itex]n = 1[/itex], then [itex](A^n)B = B(C^n)[/itex] becomes [itex](A^1)B = B(C^1)[/itex] which is[itex] AB = BC[/itex] which is supposed to be true.

Assume [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k[/itex]. Then [itex](A^k)B = B(C^k)[/itex] is true.

Prove [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k + 1[/itex].

[itex](A[/itex]k+1[itex])B = B(C[/itex]k+1[itex]) ⇒ (A(A^k))B = B(C(C^k))[/itex]

I'm not really sure how to proceed. I tried rewriting the matrix products as the summations of the products of their entries, but that didn't seem to get me anywhere. I know that I have to relate the equation when [itex]n = k + 1[/itex] to the equation when [itex]n = k[/itex] as part of induction, but I'm still stuck.

Your problem is not a lack of understanding about matrix products, etc., but a very basic lack of understanding of what *induction* is. I am going to ask you a question: do you actually know how induction is supposed to work? If you say YES, could you explain very briefly WHAT form you think such arguments take?

RGV
 
Last edited:
  • #3
No. I do not know what induction really is. My instructor only really gave one example quickly. I think he also mentioned that we should have already know what it was, but I did not learn it in AP Calculus BC.
 
  • #4
Naridax said:
No. I do not know what induction really is. My instructor only really gave one example quickly. I think he also mentioned that we should have already know what it was, but I did not learn it in AP Calculus BC.

There are numerous on-line sources that take you through some examples using induction. See,. eg., http://www.themathpage.com/aprecalc/mathematical-induction.htm or
http://zimmer.csufresno.edu/~larryc/proofs/proofs.mathinduction.html or
http://www.math.utah.edu/mathcircle/notes/induction.pdf . The first one is easiest, the second one is a bit harder and the third one is quite challenging.

RGV
 
  • #5
Hey Ray,

I'm interested in solving this problem too. If we are given the base case scenario (n=1) as true, can we use that to prove the n=k+1 case is true (in addition to using the n=k case)? The articles you linked typically avoided using the base case scenario in their proofs, but I find it difficult to use *only* the inductive assumption to prove the k+1 case. Is there some rule in induction that demands you can't use the base case, or not?
 
  • #6
Ray, I still don't know the answer to your question, but I think I know how to get the answer to the problem.

If [itex]n = 1[/itex], then [itex](A^n)B = B(C^n)[/itex] becomes [itex](A^1)B = B(C^1)[/itex] which is[itex] AB = BC[/itex] which is supposed to be true.

Assume [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k[/itex]. Then [itex](A^k)B = B(C^k)[/itex] is true.

Prove [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k+1[/itex].

[itex]A^kB = BC^k[/itex]
[itex]A(A^kB) = A(BC^k)[/itex]
[itex](AA^k)B = A(BC^k)[/itex]
[itex]A[/itex]k+1[itex]B = A(BC^k)[/itex]
[itex]A[/itex]k+1[itex]B = (AB)C^k[/itex]
[itex]A[/itex]k+1[itex]B = (BC)C^k[/itex]
[itex]A[/itex]k+1[itex]B = B(CC^k)[/itex]
[itex]A[/itex]k+1[itex]B = BC[/itex]k+1

[itex]A^kB = BC^k[/itex]
[itex](A^kB)C = (BC^k)C[/itex]
[itex](A^kB)C = B(C^kC)[/itex]
[itex](A^kB)C = BC[/itex]k+1
[itex]A^k(BC) = BC[/itex]k+1
[itex]A^k(AB) = BC[/itex]k+1
[itex](A^kA)B = BC[/itex]k+1
[itex]A[/itex]k+1[itex]B = BC[/itex]k+1

Therefore, [itex]A^nB = BC^n[/itex] is true for [itex]n = k+1[/itex], so [itex]A^nB = BC^n[/itex] is true for all [itex]n∈ℕ[/itex].

The property of matrix multiplication needed for this proof is the associative law of matrix multiplication.
 
  • #7
Naridax said:
Ray, I still don't know the answer to your question, but I think I know how to get the answer to the problem.

If [itex]n = 1[/itex], then [itex](A^n)B = B(C^n)[/itex] becomes [itex](A^1)B = B(C^1)[/itex] which is[itex] AB = BC[/itex] which is supposed to be true.

Assume [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k[/itex]. Then [itex](A^k)B = B(C^k)[/itex] is true.

Prove [itex](A^n)B = B(C^n)[/itex] is true for [itex]n = k+1[/itex].

[itex]A^kB = BC^k[/itex]
[itex]A(A^kB) = A(BC^k)[/itex]
[itex](AA^k)B = A(BC^k)[/itex]
[itex]A[/itex]k+1[itex]B = A(BC^k)[/itex]
[itex]A[/itex]k+1[itex]B = (AB)C^k[/itex]
[itex]A[/itex]k+1[itex]B = (BC)C^k[/itex]
[itex]A[/itex]k+1[itex]B = B(CC^k)[/itex]
[itex]A[/itex]k+1[itex]B = BC[/itex]k+1

[itex]A^kB = BC^k[/itex]
[itex](A^kB)C = (BC^k)C[/itex]
[itex](A^kB)C = B(C^kC)[/itex]
[itex](A^kB)C = BC[/itex]k+1
[itex]A^k(BC) = BC[/itex]k+1
[itex]A^k(AB) = BC[/itex]k+1
[itex](A^kA)B = BC[/itex]k+1
[itex]A[/itex]k+1[itex]B = BC[/itex]k+1

Therefore, [itex]A^nB = BC^n[/itex] is true for [itex]n = k+1[/itex], so [itex]A^nB = BC^n[/itex] is true for all [itex]n∈ℕ[/itex].

The property of matrix multiplication needed for this proof is the associative law of matrix multiplication.

Yes: you have it now. You could shorten the argument a lot, but you do include all the needed steps.

RGV
 
  • #8
May I ask what you meant by what "form" I think mathematical induction proofs take?
 

1. What is a matrix polynomial?

A matrix polynomial is a polynomial expression in which the coefficients and variables are matrices. It can be written in the form of cnAn + cn-1An-1 + ... + c1A + c0, where cn, cn-1, ..., c1, c0 are constant matrices and An, An-1, ..., A are matrices of the same size as the constant matrices.

2. How do you multiply two matrix polynomials?

To multiply two matrix polynomials, you can use the distributive property and the associative property of matrix multiplication. First, distribute the first matrix polynomial's terms to the second matrix polynomial. Then, use the associative property to group the terms with the same power of the variable together. Finally, use the rules of matrix multiplication to multiply the constant matrices and the variable matrices together.

3. What is the degree of a matrix polynomial?

The degree of a matrix polynomial is the highest power of the variable in the polynomial. For example, in the expression 2A3 + 5A2 + A, the degree is 3. The degree is important because it determines the size of the resulting matrix when multiplying two matrix polynomials.

4. Can you factor a matrix polynomial?

Yes, it is possible to factor a matrix polynomial into a product of simpler matrix polynomials. This is similar to factoring regular polynomials, where you look for common factors and use techniques such as grouping or the quadratic formula. The factorization of a matrix polynomial can be useful in simplifying complex expressions and solving systems of equations.

5. How is the proof of matrix polynomial multiplication different from regular polynomial multiplication?

The proof of matrix polynomial multiplication is different from regular polynomial multiplication because matrices do not follow the commutative property, meaning that A*B is not always equal to B*A. Therefore, when multiplying two matrix polynomials, the terms must be multiplied in a specific order to ensure the correct result. Additionally, the rules of matrix multiplication, such as the requirement for the number of columns in the first matrix to match the number of rows in the second matrix, must be followed throughout the proof.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
267
  • Calculus and Beyond Homework Help
Replies
2
Views
154
  • Calculus and Beyond Homework Help
Replies
1
Views
414
  • Calculus and Beyond Homework Help
Replies
3
Views
637
  • Calculus and Beyond Homework Help
Replies
3
Views
467
  • Calculus and Beyond Homework Help
Replies
1
Views
451
  • Calculus and Beyond Homework Help
Replies
1
Views
457
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
988
  • Calculus and Beyond Homework Help
Replies
2
Views
923
Back
Top