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Row of a matrix is a vector along the same degree of freedom 
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#1
Oct2909, 02:51 PM

P: 1,133

A particular introduction to matrices involved viewing them as an array/list of vectors (column vectors) in R^{n}. 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 matrixvector 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?



#2
Oct3009, 05:03 AM

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Technically, row vectors are transpose vectors, so the first row of A is is not the vector a_{1} (say), but the transpose vector a_{1}^{T}. Then your scalar product in matrix form is (a_{1}^{T})v, but in ordinary form is a.v 


#3
Oct3009, 06:00 AM

P: 1,212

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 2D plot, f(m,n) = the (m,n)th element of the matrix. 


#4
Oct3009, 09:52 AM

P: 1,133

Row of a matrix is a vector along the same degree of freedom
I see...good stuff so far...Thanks for all the replies.



#5
Nov109, 04:56 AM

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Hi Gear300!
I've found a page 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 ω^{T}L = 1/2 ω^{T}Îω where L is the (instantaneous) angular momentum, and Î is the moment of inertia tensor, both measured relative to the centre of mass. 


#6
Nov109, 08:48 AM

P: 1,133

Thanks a lot for the link tinytim.
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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