sniffnoy: (Chu-Chu Zig)
[personal profile] sniffnoy
Suppose we have a matrix representing a linear transformation. Taking its determinant is only a meaningful operation if the domain and the target space are identified and are using the same basis. But the easiest way to compute the determinant is by row reduction (or column reduction), which is only really a meaningful operation if the domain and target bases *don't* need to be the same (and one of them is considered to be fixed).

Is this just how it is ("yup, computation often requires you to do apparently meaningless things") or am I thinking about something wrong? Is there something I'm missing here?

-Harry

Date: 2012-06-07 03:26 am (UTC)
From: [identity profile] joshuazelinsky.livejournal.com
There's often disconnects between what is easy to compute. The semi-canonical example is how it is easier to formally work out the derivative of an elementary function but it is tough given an approximate set of data to work out the derivative near a given point. On the other hand, it is numerically really easy to approximate integrals, but it isn't even possible in the general case to integrate an elementary function, and it is computationally difficult to do so even when it has an elementary anti-derivative.

What I think is sort of going on in the case of the determinant is that to calculate it you are essentially going from one basis to another basis, always keeping it meaningful but simplifying the basies at each step. But that's not fully what's going on here. So I'm not sure.

Random thought with probably no usefulness at all: is this disconnect at all related to why calculating the permanent is hard?

Date: 2012-06-07 04:29 am (UTC)
From: [identity profile] sniffnoy.livejournal.com
Random thought with probably no usefulness at all: is this disconnect at all related to why calculating the permanent is hard?

I don't really see how it could be. The permanent just doesn't support row reduction. Also it's fundamentally a property of a matrix, not a linear transformation; similar matrices need not have the same permanent. These seem to be more (part of) the obstacle.

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