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


by Gear300
Tags: john baez, transpose
Gear300
Gear300 is offline
#1
Oct29-09, 02:51 PM
P: 1,133
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?
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
tiny-tim
tiny-tim is offline
#2
Oct30-09, 05:03 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,167
Quote Quote by Gear300 View Post
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!

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
How would one geometrically interpret a matrix?
Dunno , except that I always think of a matrix as being a rule that converts one vector into another vector.
mikeph
mikeph is offline
#3
Oct30-09, 06:00 AM
P: 1,197
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.

Gear300
Gear300 is offline
#4
Oct30-09, 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.
tiny-tim
tiny-tim is offline
#5
Nov1-09, 04:56 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,167
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 ωTL = 1/2 ωTω

where L is the (instantaneous) angular momentum, and is the moment of inertia tensor, both measured relative to the centre of mass.
Gear300
Gear300 is offline
#6
Nov1-09, 08:48 AM
P: 1,133
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?


Register to reply

Related Discussions
The Matrices Calculus & Beyond Homework 12
Matrices - unitary matrices Calculus & Beyond Homework 6
Matrices Calculus & Beyond Homework 3
some T/F on matrices Calculus & Beyond Homework 0