# are these matrices invertible?

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

• 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. – David Hill Dec 1 '17 at 0:24
• yes thats what i was thinking – mathew Dec 1 '17 at 0:40
• could you provide a proof that they are linearly independent? – mathew Dec 1 '17 at 0:40

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

• Thanks for giving the name of this type of matrix, now I can research it myself – mathew Dec 1 '17 at 0:45