Row of a matrix is a vector along the same degree of freedom

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Matrices can be viewed as arrays of vectors, where rows are transposed vectors, leading to the interpretation that the scalar product of a column vector and a row of a matrix involves the transpose of the row vector. This raises questions about the geometric interpretation of matrices, which some view as operators that convert one vector into another, akin to a rule for transforming functions. The discussion also touches on representing matrices geometrically through discrete plots or transformations in function space. Additionally, the relationship between ordinary and matrix representations is highlighted, particularly in the context of angular momentum and kinetic energy equations. Overall, the conversation explores the complexities of matrix representation and their geometric implications.
Gear300
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A particular introduction to matrices involved viewing them as an array/list of vectors (column vectors) in Rn. The problem I see in this is that it is sort of like saying that a row of a matrix is a vector along the same degree of freedom (elements of the same row are elements of different vectors all in the same dimension). So from this, technically, the scalar product of a column vector v and row1 of a matrix A should only exist as a product between the elements of row1 of the matrix A and row1 of the column vector v...which doesn't seem right (since matrix-vector multiplication Av is defined as a column vector of dot products between the vector v and rows of A). How would one geometrically interpret a matrix?
 
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Gear300 said:
… So from this, technically, the scalar product of a column vector v and row1 of a matrix A should only exist as a product between the elements of row1 of the matrix A and row1 of the column vector v...

which doesn't seem right (since matrix-vector multiplication Av is defined as a column vector of dot products between the vector v and rows of A).

Hi Gear300! :smile:

Technically, row vectors are transpose vectors,

so the first row of A is is not the vector a1 (say), but the transpose vector a1T.

Then your scalar product in matrix form is (a1T)v,

but in ordinary form is a.v :wink:
How would one geometrically interpret a matrix?

Dunno :redface:, except that I always think of a matrix as being a rule that converts one vector into another vector. :smile:
 


I understand matrices as operators, like you may have basis vectors (x,y,z all normal), you can have basis functions (such as sin(nx) for n=1,2,3...) which any periodic function can be decomposed into using Fourier series. Then the function can be represented completely by a column vector containing the amplitude of each frequency, and the matrix multiplication will output a new function. So in that sense you might represent a matrix geometrically by a series of "before and after" functions!

Then again this is just fun speculation... the most fundamental way of expressing a matrix geometrically is probably by a discrete 2-D plot, f(m,n) = the (m,n)th element of the matrix.
 


I see...good stuff so far...Thanks for all the replies.
 
Hi Gear300! :smile:

I've found http://farside.ph.utexas.edu/teaching/336k/lectures/node78.html#mom" on john baez's excellent website which shows the relation between the ordinary and the matrix representation of the same equation, and where the transpose fits in …

if ω is the (instantaneous) angular velocity, then the rotational kinetic energy is both:

1/2 ω.L
and (without the "dot")
1/2 ωTL = 1/2 ωTÎω

where L is the (instantaneous) angular momentum, and Î is the moment of inertia tensor, both measured relative to the centre of mass.
 
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Thanks a lot for the link tiny-tim.

So, its sort of like saying that for an m x p matrix with n being a dimensional space, you could label the rows (going down) from n = 1 to n = m and you could also label the columns (going across) from n = 1 to n = p, right?
 
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