View Full Version : Singular Matrices
Antonio Lao
Mar9-04, 09:23 AM
I am getting all singular matrices in my research on the quantization of one-dimensional space.
I think this is the proper place for me to ask all mathematicians the disadvantage of singular matrix.
I know that singular matrix does not have an inverse since its determinant is zero and hence a metric cannot be defined.
What is the parctical use of a metric besides giving a "distance?"
Zero metric can also mean zero mass in physics. But if all masses are zeros, where do the experimentally determined masses come from?
matt grime
Mar9-04, 09:45 AM
Matrices do have a metric (many in fact) irrespective of whether or not they are singular, the only time d(M,N) is zero if d is a matrix metric is if M=N
metrics yield topologies, but as a metric is a (generalized) notion of 'distance', I'm not sure I understand what you're asking.
singular of a matrix means that the codimension of the image is not zero, that do you for another interpretation? or the dimension of the kernel is non-zero.
anyway, just becuase the determinant is not a metric does not imply there is no metric in the space of matrices. There is an impotant one, and it implies that if M is an invertible matrix, and H is a matrix such that d(M,M+H) < 1/d(M,0) that M+H is invertible too.
Antonio Lao
Mar9-04, 12:19 PM
matt grime,
Thanks for helping me again with my math deficit.
Are the following square symmetric matrices singular and do they have metric, inverse, etc.?
H^{+} = \left(\begin{array}{cc}+1 & -1\\-1 & +1\end{array}\right)
and
H^{-} = \left(\begin{array}{cc}-1 & +1\\+1 & -1\end{array}\right)
Isn't nonzero determinant a necessary condition for having a metric?
Please clarify further?
Antonio
matt grime
Mar9-04, 12:35 PM
both those are singular matrices yes, but there are still metrics on (2x2) matrices and the metric is not zero on them.
let [a,b,c,d] denote the matrix
|a b|
|c d|
then one norm is
n([a,b,c,d]) = sqrt(|a|^2 + |b|^2 + |c|^2 + |b|^2)
another is the usual operator norm:
n(M) = sup ||M(x)||
where sup is taken over all vectors x of norm 1.
A metric can be defined from a norm:
d(A,B) = n(A-B)
neither matrix has an inverse, but that isn't important really.
nxm matrices are just nxm dimensional real space, where there are norms too.
det isn't a metric - det can be -ve, for instance.
perhaps you should say what you think a metric is?
Antonio Lao
Mar9-04, 04:50 PM
matt grime,
Thanks for your continued elucidations.
I am still having difficulty understanding your math notations and their meanings.
The truth is I don't know what a metric is. At first I thought metric defines a "distance" in some space.
This might be a stupid question, can there be a metric in one-dimensional space, in two-dim space, in 3-dim space?
The important question for me is how do we define a metric in one-dim space using a matrix?
Antonio
selfAdjoint
Mar9-04, 05:15 PM
If X is some space, a metric is a map d: X \times X \rightarrow R that associates each pair of points in X to a real number, that satisfies
d(x,x) = 0
d(x,y) = d(y,x); and
d(x,z) \le d(x,y) + d(y,z)
As long as they obey these laws, metrics don't have to be nice, and there are metric spaces that don't have anything to do with euclid.
Antonio Lao
Mar9-04, 07:10 PM
selfAdjoint,
Your notations are a lot better than matt's but still I'm having troubles with your math logic.
Is d:X x X -->R a one-dim metric? What is the operator x? Is it multiplication?
How to write the map for a 2-dim metric, or a 3-dim metric?
There is a metric in GR where the g11 is 1, g22=-1, g33=-1, g44=-1 and all the other elements are zeros. What is the meaning or purpose of this metric in GR? (another stupid question).
Antonio
matt grime
Mar9-04, 07:40 PM
the small x there means cartesian product.
simply what is being got at is that the distance notion requires two inputs - the initial point and the final point; the dsitance is from a point to another point.
I'm sorry that you find my notations unclear, but they are very standard.
you ought to know what a norm is and what sup means; they are easily learned ideas; wolfram can help, usually.
a metric satisfies 3 rules:
two points are zero distance apart iff they are the same point
the distance from a to b is the same as the distance from b to a,
the distance from a to b directly is shorter than going via nay intermediate point.
How does one define a metric in 1d using a matirx? dunno, because the only matrices from R to R are 1x1, ie R again, and the only singular 1x1 matix is 0... and i've no idea what you are getting at.
i though you wanted a metric on some rxp matrices. of which there are infintely many.
what do you mean by an n-dim metric? metrics do not have a dimension.
i really don't understand what's wrong with my notation; in fact i'm more than slightly offended by the implication. perhaps you should learn what the meanings of the terms you presumne to use are first?
selfAdjoint
Mar10-04, 10:01 AM
Originally posted by Antonio Lao
selfAdjoint,
Your notations are a lot better than matt's but still I'm having troubles with your math logic.
Is d:X x X -->R a one-dim metric? What is the operator x? Is it multiplication?
How to write the map for a 2-dim metric, or a 3-dim metric?
There is a metric in GR where the g11 is 1, g22=-1, g33=-1, g44=-1 and all the other elements are zeros. What is the meaning or purpose of this metric in GR? (another stupid question).
Antonio
g^{\mu \nu} is the metric tensor. The metric between two points u_{\mu} and v_{\nu} is g^{\mu \nu}u_{\mu}v_{\nu} where the Einstein summation condition is observed, i.e. you add up all the terms for \mu, \nu = 0, 1, 2, 3 . This gives you a unique real number. You can verify for your self with examples that this definition of a metric - a real number associated with each pair of points - satisfies the three metric axioms I stated.
Antonio Lao
Mar10-04, 12:21 PM
matt and selfAdjoint,
Thanks to both of you for the valuable helps. These will help me save time but still I have to go on WolframResearch - Eric Weisstein's world of Math and Physics.
At least now I know what to look for. Thanks again.
Antonio
Antonio Lao
Mar10-04, 12:41 PM
matt grime,
you quote
_________________
i really don't understand what's wrong with my notation; in fact i'm more than slightly offended by the implication. perhaps you should learn what the meanings of the terms you presumne to use are first?
__________________
There is nothing wrong with your notations, I just can't understand them. The faults is mine. I am trying to learn the meanings and the notations simultaneously and to me it does present a problem. If I stumble on Egyptians hieroglyphics, I would have the same problem. I know there is nothing wrong with the hieroglyphics, the message is there but it is in codes.
Antonio
matt grime
Mar10-04, 12:51 PM
ok,my reply was far too nasty in tone anyway and i was going to take it down, or at least edit it.
Antonio Lao
Mar10-04, 12:57 PM
matt grime,
It's too late for that, the whole world (the entire universe) saw it.
Antonio
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