Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was given a theorem:

The polynomials (where $f$ and $g$ are complex polynomials of degrees $n$ and $m$)

$$f(z), zf(z), \ldots , z^{m−1}f(z), g(z), zg(z), \ldots,z^{n−1}g(z)\tag{7.6.4}$$

form a basis of the space $P(\mathbb{C},m + n − 1)$ if and only if $f$ and $g$ have no common zero.

And also a problem:

Find a basis, and the dimension, of the space of polynomials spanned by the polynomials in $(7.6.4)$ when $f(z) = z^2(z − 1)$ and $g(z) = z^2(z + 1)$

Now, obviously, $f$ and $g$ have a common zero, namely - $0$. So I cannot use the theorem, can I? If no, what the other way around it?

share|improve this question
    
What is $P(\mathbb{C},m+n-1)$? –  Dedalus Jun 16 '13 at 18:59
    
Yeah, sorry! This is the space of complex polynomials of degree at most $n+m-1$ –  Sarunas Jun 16 '13 at 19:01
    
In that case - you can't use the theorem above no, they have a common zero. But you don't have to use this to compute the dimension or find a basis of the space spanned by the polynomials above! –  Dedalus Jun 16 '13 at 19:05
    
I have a question: is the dimension for this space 2? As if indeed it is, the problem is easier than I think. if not, I'm kind of lost :) –  Sarunas Jun 16 '13 at 19:10
    
You have $m = n = 3$. Put that and $f$ and $g$ in $(7.6.4)$. The point of the problem is probably to show you that some of the obtained set of polynomials is not independent. Eliminate those and you have your basis (and a dimension). –  Vedran Šego Jun 16 '13 at 21:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.