Math Is Fun Forum

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Matrix

The asscociative law is (AB)C=A(BC) , what happen if more than three matrix , ABCD=A(BCD)=A(BC)D????

Last edited by Stanley_Marsh (2007-04-23 09:35:16)

---

Numbers are the essence of the Universe

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Matrix

i've never understood why this is so in the first place, i would have though it simple to infer that since AB =/= BA, you would have automaticly that (AB)C =/= A(BC), but ofcourse, its not the case.

ABCD=A(BCD)=A(BC)D????

(AB)CD = A(BC)D = AB(CD)

thats the way i see it, and testing with 4 random matrices, it holds.

Last edited by luca-deltodesco (2007-04-23 09:46:30)

---

The Beginning Of All Things To End.
The End Of All Things To Come.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Matrix

There is a theorem that says that if an operation is associative and more than two elements are combined, then all the different ways of putting brackets around the elements to evaluate the product will yield the same result. The order of the elements have to be the same, of course (unless the operation is commutative).

Last edited by JaneFairfax (2007-04-23 21:18:21)

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Matrix

i've never understood why this is so in the first place, i would have though it simple to infer that since AB =/= BA, you would have automaticly that (AB)C =/= A(BC), but ofcourse, its not the case.

What does the noncommutativity of matrix multiplication have to do with the associativity? BA doesn't show up in (AB)C = A(BC).

The definition of associativity is that the order of operations does not matter, as long as the order of operands doesn't change. From this it follows easily that (AB)CD = A(BC)D = AB(CD) = (ABC)D = A(BCD); we can multiply in any order (A and B first, or B and C first, etc.), but we cannot reorder the operands (A, B, C, and D).

If you really don't believe it, take the product of r matrices A[sub]k[/sub] (where each A[sub]k[/sub] is of dimension n[sub]k[/sub] × n[sub]k + 1[/sub]), denoting the elements of A[sub]k[/sub] as

, and write out what it would be:

If we take, for example,

, this sum becomes

by the associativity of scalar multiplication (which I am sure you do not dispute), and similarly for other placings of parentheses (they will all reduce to parentheses around the scalar elements in the sum, which can be moved as removed as you please).

Last edited by Zhylliolom (2007-04-23 11:22:23)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Matrix

I've got it , thanks .

---

Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Matrix

One more question , if A , B have inverse ,so does A+B?

---

Numbers are the essence of the Universe

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Matrix

No. Consider

and

They are both invertible (they are their own inverses), but I + (-I) = O, and we know the zero matrix is not invertible. Kind of a simple example, but it works.

Last edited by Zhylliolom (2007-04-23 20:45:17)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Matrix

A has inverse , B has inverse , is there a proof for A+B?

---

Numbers are the essence of the Universe

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Matrix

No proof for A+B- just think of I and -I.

A(BCD)=A((BC)D)=A(B(CD))=ABCD= (ABC)D= (A(BC))D= A(BC)D

---

X'(y-Xβ)=0

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Matrix

There is a theorem that says that if an operation is associative and more than two elements are combined, then all the different ways of putting brackets around the elements to evaluate the product will yield the same result. The order of the elements have to be the same, of course (unless the operation is commutative).

Here is a proof by strong induction.

Let • be an associative operation and consider the product

.

If n = 1 or 2, there is only one way to evaluate r. If n = 3, there are two ways to do it, and they are equal by the associativity of •.

Let n > 3 and suppose the product can be uniquely evaluated for all k < n. Then either

or

By the inductive hypothesis,

  and

  can be uniquely evaluated; moreover,

Also, since • is associative,

Hence

That completes the proof (which came to me in a dream last night).

Last edited by JaneFairfax (2007-04-25 23:29:31)