Please consider the following matrix
$$M_{m+1\times m+1}= \begin{bmatrix} \frac{1}{2m+1} & \frac{1}{2m} & \dots & \frac{1}{m+2} & \frac{1}{m+1} \\ \frac{1}{2m} & \frac{1}{2m-1} & \dots & \frac{1}{m+1} & \frac{1}{m} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ \frac{1}{m+2} & \frac{1}{m+1} & \dots & \frac{1}{3} & \frac{1}{2}\\ \frac{1}{m+1} & \frac{1}{m} & \dots & \frac{1}{2} & 1 \end{bmatrix} $$
I have faced this matrix while proposing an approximate solution for a dynamic system. Proving invertibility of $M$ equals proving solve-ability of the dynamic system.
- This matrix is symmetric but this property does not seem such helpful for proving invertibility.
- The denominator of this matrix is added by one when moving one cell toward left or above, but it also does not seem very helpful for proving existence of determinant.
- I tried to use Matlab for computing the determinants and induce a rule from them. However, for $m>5$ Matlab does not compute it exactly (i.e. $\text{inv}(M)\cdot M\neq I_{m+1}$).
May anyone provide any hint for proving invertibility of $M$?
Let $\phi_j(x)=x^{m+1-j}$, $j=0,\ldots,m$. Then your matrix $M$ is the Gram matrix of the elements $\phi_0,\ldots,\phi_m$ with respect to the positive definite inner product $$\left<f,g\right>=\int_0^1 f(x)g(x)\,dx$$ on the space of continuous functions from $[0,1]$ to $\Bbb R$. As the $\phi_j$ are linearly independent, then $M$ is positive definite.