Dear friends,
I have a problem in matrix theory, I want to prove that the inverse of the following matrix exist.
Consider ai , j=1,2,...n, belong to the set of real number, with aj are in assending order and if we have matrix
A=[D{i,j}]
where D{i,j}=ai j,
The inverse of the matrix exist as I find its inverse with MATLAB.
Dear Hammad, from your attachment above, the matrix A is invertible if, and only if
det A \ne 0. Now by the determinant properties, the determinant of the given matrix
det A = (\tau_1)...(\tau_n) det V,
where det V is the well-known determinant of Vandermonde matrix, it is not equal to 0 if \tau_j \ne 0, j=1,2, ..., n.
Dear @Hammad, I do not see that the matrix should be invertible; perhaps it is not easy to show that it is singular but it is not clear why should it be non-singular.
Dear Hammad, from your attachment above, the matrix A is invertible if, and only if
det A \ne 0. Now by the determinant properties, the determinant of the given matrix
det A = (\tau_1)...(\tau_n) det V,
where det V is the well-known determinant of Vandermonde matrix, it is not equal to 0 if \tau_j \ne 0, j=1,2, ..., n.
Then the inverse exist as pointed out by wiwat that the determinant of this matrix is a product of \tau_i with a Vandermonde matrix.
Also you don't need the condition that they are increasing as long as they are non-zero and pairwise distinct, determinant will be non-zero.
Dear Sir,
This is a Vandermonde matrix (which is always full rank), unless you have a row (or column) which is all zeros, or two rows (or columns) that are generated by the same number, in other words: if all your a_i are different and non zero, then the inverse exists.
For the proof, refer to the following:
http://math.stackexchange.com/questions/426932/why-are-vandermonde-matrices-invertible
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