I wasn't sure if should be immediately obvious to me that the Bernstein basis (for n) is a basis for polynomials of degree n. I looked it up and it seems nontrivial. See pages 8-10 here: http://www.idav.ucdavis.edu/education/CAGDNotes/Bernstein-Polynomials.pdf
keenan
Try doing it this way: you can expand the Bernstein bases in the canonical basis $1, x, x^2, ...$. If the coefficients in this basis are linearly independent, then you know that the Bernstein bases are linearly independent.
I wasn't sure if should be immediately obvious to me that the Bernstein basis (for n) is a basis for polynomials of degree n. I looked it up and it seems nontrivial. See pages 8-10 here: http://www.idav.ucdavis.edu/education/CAGDNotes/Bernstein-Polynomials.pdf
Try doing it this way: you can expand the Bernstein bases in the canonical basis $1, x, x^2, ...$. If the coefficients in this basis are linearly independent, then you know that the Bernstein bases are linearly independent.