Matrix Dot Product: Solving Linear Algebra

nigels
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Hi, here's the problem:

for m = {{a, b}, {c, d}},

m \cdot m is suppose to = {{a^2 + b c, a b + b d}, {a c + c d, b c + d^2}}

It's been ages since I took linear algebra and now can't figure out how this works.

Thanks for your help!
 
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That is the product M*M=M^2. When you multiply two matrices, the product will have in its ith row and jth column the dot product of the ith row of the left matrix times the jth column of the right matrix. So to get the first entry (entry in the first row, first column), you perform the dot product of (a, b) -- the first row of the first matrix with (a,c)--the first column of the second matrix. That is a*a+b*c=a^2+bc.
 
Is M \cdot M the same thing as M*M?
 
nigels said:
Is M \cdot M the same thing as M*M?

It depends on how the 'dot' has been defined. In this case it is the product of M with itself so i suspect it just gave the formula for M^2. However, it is not a "dot product". Dot products -- more generally inner products -- give back a number.
 
Hmm. I thought so.. that dot product works by summing the product of individual elements of vectors, which is why it's confusing when Mathematica uses the notation "." in the context of matrix multiplication. So, if I understand correctly, one can't really take the dot product of matrices and what I see is only the result of notational configuration by the computing language.
 
nigels said:
Hmm. I thought so.. that dot product works by summing the product of individual elements of vectors, which is why it's confusing when Mathematica uses the notation "." in the context of matrix multiplication. So, if I understand correctly, one can't really take the dot product of matrices and what I see is only the result of notational configuration by the computing language.

I would have to agree with you. Though, you can define an inner product on m x n matrices by adding the products of the corresponding entries in the matrices. It ends up being the same as the standard dot product on vectors in R^mn.
 
Thanks for the help, Davey.
 
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