thanks for the help!

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- Jan 17th 2010, 10:05 AMMMAfanaticDimension of span(Z)
thanks for the help!

- Jan 17th 2010, 11:36 AMDinkydoe
First: You can write these polynomials with respect to the basis

Thus we can write:

Put these vectors in a matrix and by performing elementary row-operations we get:

We see that**span**is a 3-dimensional subset and:

Now, letting:

:

is such that**span**( )=**span**( ) - Jan 17th 2010, 12:24 PMMMAfanatic
hey thank you for your help.

could u please send the operations you used to row reduce the matrix please.

when i did it i just did row 2 - 2 x row 1, row 3 - 4 x row 1, row 4 - row 1, row 5 - 13 x row 1, row 3 + 3x row 2

so im left with

1 0 1 1 0

0 1 0 -2 1

0 -3 -1 0 0

0 -1 -2 -10 5

i know that i am wrong because im not sure how to row reduce properly i dont think :-(

thanks for you help again. - Jan 17th 2010, 01:18 PMDinkydoe
Actually I let Maple calculate the reduced form. I admit it can be quite tedious work, but it's still probably the simplest method.

- Jan 17th 2010, 01:40 PMMMAfanatic
oh right...fair enough lol.

umm i know row 2 is row 2 - 2 x row 1, dont know how the other rows came about?

can anyone help on the row reducing to get the final product.

sorry guys for this.

major thanks to the original helper and anyone else who may be able to help me out :-) - Jan 17th 2010, 01:53 PMMMAfanatic
i figured it

:-)

[/COLOR] - Jan 17th 2010, 02:00 PMDinkydoe
Very good. Now find even 2 more linear independant vectors: , write them on the basis to find polynomials

Then together with you have a basis for - Jan 17th 2010, 03:53 PMMMAfanatic
- Jan 17th 2010, 04:11 PMDinkydoe
No, it's the part where they ask to extend to find a basis for .

But they can be used to extend Y.

Namely is independant of Y and is independant of Y

We can even show that: is a basis for

You can do this by showing that:

has rank 5. - Jan 17th 2010, 04:36 PMMMAfanatic
- Jan 17th 2010, 05:23 PMDinkydoe
It's the same matrix with the 3 upper rows:

with and instead of the zero-rows.

Where are the vector representations of w.r.t. the basis

You want to show that is linear independant of . You can do that by showing that are linear independant, that is, the given matrix has rank 5. (meaning, having 5 linear independant rows). - Jan 17th 2010, 06:49 PMMMAfanatic
oh i see.

i know why i was confused. i disagree with what your final row operated product looks like

i have

1 0 1 1 0

0 1 0 -2 1

0 0 1 6 -3

0 0 0 0 0

0 0 0 0 0 - Jan 17th 2010, 06:55 PMDinkydoe
You're allways free to disagree with a computer output

I didn't do the calculations, do whatever suits you.

edit: Actually they're the same matrices.