Krdytkyu

Hello I want to show that the Bernstein polynomial $$B_{n,k}=binom{n}{k}x^k(1-x)^{n-k},$$ is a basis. For linear independece I got a hint from my teacher to expand the binom $(1-x)^{n-k}$ This way I get: $$B_{n,k}=binom{n}{k}x^k(1-x)^{n-k}sum_{j=0}^{n-k}binom{n-k}{j}(-1)^jx^j$$ And changing the index of summation gives: $$B_{n,k}=sum_{j=k}^{n}binom{n-k}{j-k}binom{n}{k}(-1)^{j-k}x^{j-k+k}=sum_{j=k}^{n}binom{n}{j}binom{j}{k}(-1)^{j-k}x^j$$ Now I have to show that $alpha_i$ are $0$ in the relation $sum_{i=0}^{n}alpha_iB_{i,n}=0,$ or$$alpha_0sum_{j=0}^n(-1)^jbinom{n}{j}binom{j}{0}x^j+alpha_1sum_{j=1}^n(-1)^{j-1}binom{n}{j}binom{j}{1}x^j+...+alpha_{n}sum_{j=n}^n(-1)^{j-n}binom{n}{j}binom{j}{n}x^j=0$$ Now what can I do and how can I finish this problem? Thanks in advance!

When you expand them, you see that only one of the Bernstein polynomials has a non-zero constant term, namely $binom nn(1-x)^n$. So, if$$sum_{k=0}^nalpha_kB_{n,k}(x)=0,tag1$$then $alpha_0=0$.

Now, there are only two Bernstein polynomials such that the coefficient of $x$ is non-zero, which are $B_{n,0}(x)$ and $B_{n,1}(x)$. But you already know that $alpha_0=0$. It follows then from $(1)$ that $alpha_1=0$.

And so on…

Hint: The matrix that expresses the Bernstein polynomials with respect to the canonical monomial basis is triangular with a diagonal of binomial coefficients, and so is invertible.

For instance, when $n=3$, we have
$$
begin{pmatrix}
B_{3,0}(x) \ B_{3,1}(x)\ B_{3,2}(x) \ B_{3,3}(x)
end{pmatrix}
=
begin{pmatrix}
(1-x)^3 \ 3x(1-x)^2 \ 3x^2(1-x) \ x^3
end{pmatrix}
=
begin{pmatrix}
1 & -3 & hphantom{-}3 & hphantom{-}1 \
0 & hphantom{-}3 & -6 & hphantom{-}3 \
0 & hphantom{-}0 & hphantom{-}3 & -3 \
0 & hphantom{-}0 & hphantom{-}0 & hphantom{-}1 \
end{pmatrix}
begin{pmatrix}
1 \ x \ x^2 \ x^3
end{pmatrix}
$$
The exact entries in the matrix are not important. The key point is that the $x^k$ factor in $B_{n,k}(x)$ ensures that in the $k$-th row all entries before the diagonal are zero, and so the matrix is triangular.

First of all, note that we have
power series and Bernstein polynomials

Also if there exist constants $c_0, c_1, ..., c_n$ so that the identity
$0 = c_0*B_{0,n}(t) + c_1*B_{1,n}(t) + · · · + c_n*B_{n,n}(t)$
holds for all t, then all the $c_i$’s must be zero and we have
Indecency
Since the power basis is a linearly independent set, we must have that
Zero coefficient
which implies that $c_0 = c_1 = · · · = c_n = 0$ ($c_0$ is clearly zero, substituting this in the second equation gives $c_1 = 0$, substituting these two into the third equation gives ...)