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Wednesday, August 8, 2007

Just surrender and it won't hurt at all

Fixed questions 1-3. Proving that z is unique was particularly pleasing, though quite simple. My use of the implicit function theorem to do this initially was incorrect; turns out the Im. F. T. is only for regions very close to a particular point, not for the entire domain. My final proof doesn't require anything very advanced, which is a sure sign that it's on the right track.

My next challenge is to show that the determinant of a matrix is a continuous, infinitely differentiable function [question 7]. That it's a function is easy to see - clearly each matrix has exactly one determinant; proving that it's continuous has me a little bewildered. Doing so will mean that every possible determinant has at least one associated matrix. That's a pretty incredible thought, really.

The great thing, for me, about studying math is that it's very easy to lose yourself in when you want to forget about things. Everyone's gotta have their own means of escape.


gliderguider said...

hammer to fall.

Seems to me it's pretty easy to construct a matrix with any determinant you want. If you multiply every element in the matrix by X then doesn't the determinant also get multiplied by X?

Gael said...

Star for you.

If you multiply every element in the matrix by X then doesn't the determinant also get multiplied by X?

Not sure - it might be nX, where n is the number of rows - the determinant is linear in the first row, and thus in every row. And of course, yes, it's easy to construct a matrix with any determinant, but I'd never thought about it like that before!

gliderguider said...

d'oh .. it'll be n^X of course. I blame the remnants of morphine.

And I think you should be able to take an arbitray matrix and make the determinant anything you want by adjusting just any one element in a linear fashion, provided that the determinant of the matrix obtained by deleting the row and column containing that element is nonzero.

Gael said...

Well, yes. Of course.

Why n^X?

gliderguider said...

because the determinant is the sum/difference of a whole lot of terms, with each term being the product of n numbers.

lorne said...

Um, perhaps it'd be x^n? That is, n multiplications of your coefficient. Once for each row.

How is yonder shoulder?

gliderguider said...

damn. They were right about not operating heavy mathematics or making financial decisions.

I'll quit while I'm behind.

Shoulder is a PITA. I'll live. It's not the first time, just the first time in the 00's.

Gael said...

PITA? What's your shoulder doing there?

gliderguider said...

Displaying more flexibility than it was designed to.

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