0
$\begingroup$

Let X = {x0,x1,x2...xn} be a set of distinct reals, consider the vectors
(for simplicity let 0^0=1)

v0 = (x0^0,x0^1,...x0^n)
v1 = (x1^0,x1^1,...x1^n)
v2 = (x2^0,x2^1,...x2^n)
...
vn = (xn^0,xn^1,...xn^n)

Is the set {v1,...vn} linearly independent for all sets X? If so please give a proof, otherwise please give a counter example.

If they are not independent what are some conditions I could place on X to make sure they are always independent.

Also, is there a nice formula for the inverse of this matrix? If not, is there a nice formula for the inverse of the matrix you get when using a specific set X?

Thanks

$\endgroup$
  • 2
    $\begingroup$ The vectors are linearly independent. However, you have $n$ vectors in $\mathbb{R}^{n+1}$, so the matrix you get with rows $v_1,\ldots,v_n$ is not square. Maybe you would like to add another real number $x_0$? In that case, you get the Vandermonde matrix. $\endgroup$ – David Hill Dec 1 '17 at 0:24
  • $\begingroup$ yes thats what i was thinking $\endgroup$ – mathew Dec 1 '17 at 0:40
  • $\begingroup$ could you provide a proof that they are linearly independent? $\endgroup$ – mathew Dec 1 '17 at 0:40
1
$\begingroup$

The vectors are linearly independent. Why are Vandermonde matrices invertible?

However, the matrix $V$ is not square. If you add a real number $x_{0}$ you get the Vandermonde matrix, in which you can find the matrix in Matlab using the algorithm discovered by Shaohong Yan and Aimin Yang (better than MatLab for $n >= 7$):

function [DD]=invvander(a)
for i=l:n
    Y(i,l)=Tl+a(i);
    Y(i,i)=T2*a(i);
    Tl=Y(i,l);
    T2=Y(i,i);
end
for i=3:n
    for j=2:i-l
    Y(ij)=Y(i-l j)+a(i)*Y(i-l j-l);
    end
end
Y=[sym(ones(n,l)),Y];
for i=l:n
    for j=l:n
        AAG,i)=(n+ I-j)*(-a(i)Y'(n-j);
    end
end
FA=simple(Y(end,l:end-l)*AA);
for i=l:n
    for j=2:n+l
        DG,i)=(DG-l,i)-Y(end,n-j+3)/FA(i))/(-a(i));
    end
end

ref: http://ieeexplore.ieee.org/document/5413083/

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for giving the name of this type of matrix, now I can research it myself $\endgroup$ – mathew Dec 1 '17 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.