Tuesday, April 17, 2007

Properties of Matrix Multiplication

Multiplication can only occur between matrices A and B if the number of columns in A match the number of rows in B. This presents the very important idea that while multiplication of A with B might be a perfectly good operation; this does not guarantee that multiplication of B with A is a perfectly good operation.

Even if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. In other words, unlike the integers, matrices are noncommutative.

Property 1: Associative Property of Multiplication

A(BC) = (AB)C

where A,B, and C are matrices of scalar values.


(1) Let D = AB, G = BC
(2) Let F = (AB)C = DC
(3) Let H = A(BC) = AG
(4) Using Definition 1, here, we have for each D,F,G,H:

di,j = ∑k ai,k*bk,j

gi,j = ∑k bi,k*ck,j

fi,j = ∑k di,k*ck,j

(5) So, expanding fi,j gives us:

fi,j = ∑k (∑l ai,l*bl,j)ck,j =

(∑kl) ai,l*bl,k*ck,j =

= ∑l ai,l*(∑k bl,k*ck,j) =

= ∑l ai,l*gl,j = hi,j


Property 2: Distributive Property of Multiplication

A(B + C) = AB + AC
(A + B)C = AC + BC

where A,B,C are matrices of scalar values.


(1) Let D = AB such that for each:

di,j = ∑k ai,k*bk,j

(2) Let E = AC such that for each:

ei,j = ∑k ai,k*ck,j

(3) Let F = D + E = AB + AC such that for each:

fi,j = ∑k ai,k*bk,j+ai,k*ck,j = ∑k ai,k[bk,j + ck,j]

(4) Let G = B+C such that for each:

gi,j = bi,j + ci,j

(5) Let H = A(B+C) = AG such that for each:

hi,j = ∑k ai,k*gk,j

(6) Then we have AB + AC = A(B+C) since for each:

hi,j = ∑k ai,k[bk,j + ck,j]

(7) Let M = A + B such that for each:

mi,j = ai,j + bi,j

(8) Let N = (A+B)C = MC such that:

ni,j = ∑k mi,k*ck,j =

= ∑k (ai,k + bi,k)*ck,j

(9) Let O = BC such that:

oi,j = ∑k bi,k*ck,j

(10) Let P = AC + BC = E + O such that:

pi,j = ei,j + oi,j =

= ∑k ai,k*ck,j + ∑k bi,k*ck,j =

= ∑k [ai,k*ck,j + bi,k*ck,j] =

= ∑k (ai,k + bi,k)*ck,j


Property 3: Scalar Multiplication

c(AB) = (cA)B = A(cB)


(1) Let D = AB such that:

di,j = ∑k ai,k*bk,j

(2) Let E = c(AB) = cD such that for each:

ei,j = c*di,j = c*∑k ai,k*bk,j

(3) Let F = (cA)B such that:

fi,j = ∑k (c*ai,k)*bk,j = c*∑k ai,k*bk,j

(4) Let G = A(cB) such that:

gi,j = ∑k ai,k*(c*bk,j) =

= c*∑k ai,k*bk,j


Property 4: Muliplication of Matrices is not Commutative

AB does not have to = BA


(1) Let A =

(2) Let B =

(3) AB =

(4) BA =




Eva Acosta said...

I am a math teacher.
You can revise your understanding of matrices solving exercises about addition, subtraction and multiplication of matrices. Inverse,... at www.emathematics.net

Anonymous said...

i am a student taking a course in linear algebra. i just wanted to say that your tutorrial is very helpfull. so post some more on this topic.
thank you

Luis said...

Found it helpful. Thank you!

Amittai Aviram said...

Hi! Thank you very much for providing this resource! I am troubled, however, by something in your proof of the associativity of matrix multiplication. (Since blogger.com will not let me use the necessary HTML tags, I will attempt to use LaTeX-like notation instead.)

In (4), you have
d_{i,j} = \sum_{k} a_{i,k}*b_{k,j}
f_{i,k} = \sum_{k} d_{i,k}*c_{k,j}
Then, in (5), you expand this last equation as
f_{i,j} = \sum_{k} (\sum_{l} a_{i,l}*b_{l,j}) c_{k,j}
However, we are expanding d_{i,k} from the previous equation, not d_{i,j} (where the outer summation symbol binds the variable k). And
d_{i,k} = \sum_{l} a_{i,l}*b_{l,k}
The summation symbol that binds k becomes the outer summation symbol in the expansion. So the correct expansion of the whole equation should be:
f_{i,j} = \sum_{k} (\sum_{l} a_{i,l}*b_{l,k}) c_{k,j}
which does not lend itself to so neat a proof as the one you present. A correct proof, though with rather messier notation, can be found here:

Amittai Aviram said...

Another page that has a correct proof (avoiding the error noted above) can be found here:

Larry Freeman said...

Hi Amittai,

Thanks very much for pointing out the mistake. I will revise the proof and post a comment on this blog when it is fixed.