Let \[ H_{ij}=\int_0^1t^it^j\,dt,\qquad0\le i,j\le n, \] and let \(P_i(t)=\sum_{j=0}^ip_{ij}t^j\) be orthonormal on \([0,1]\). If \(P=(p_{ij})\), show that \[ H^{-1}=P^TP. \]
Solution
Extend the coefficients by setting \[ p_{ij}=0 \] whenever \(j>i\), so that \(P=(p_{ij})_{0\le i,j\le n}\) is an \((n+1)\times(n+1)\) matrix.
1. For every \(0\le i,j\le n\),\[ (PHP^T)_{ij} = \int_0^1P_i(t)P_j(t)\,dt. \]
By matrix multiplication, \[ \begin{aligned} (PHP^T)_{ij} &= \sum_{\alpha=0}^n\sum_{\beta=0}^n p_{i\alpha}H_{\alpha\beta}p_{j\beta}\\ &= \sum_{\alpha=0}^n\sum_{\beta=0}^n p_{i\alpha}p_{j\beta} \int_0^1t^\alpha t^\beta\,dt. \end{aligned} \] Because these are finite sums, they may be moved inside the integral: \[ \begin{aligned} (PHP^T)_{ij} &= \int_0^1 \left(\sum_{\alpha=0}^np_{i\alpha}t^\alpha\right) \left(\sum_{\beta=0}^np_{j\beta}t^\beta\right) \,dt\\ &= \int_0^1P_i(t)P_j(t)\,dt. \end{aligned} \]
2. The orthonormality hypothesis gives\[ \boxed{PHP^T=I}. \]
By step 1 and the assumed orthonormality, \[ (PHP^T)_{ij} = \int_0^1P_i(t)P_j(t)\,dt = \delta_{ij}. \] Thus every entry of \(PHP^T\) agrees with the corresponding entry of the identity matrix.
3. The matrix \(P\) is invertible.
From step 2, \[ n+1 = \operatorname{rank}(I) = \operatorname{rank}(PHP^T) \le \operatorname{rank}(P) \le n+1. \] Hence \[ \operatorname{rank}(P)=n+1, \] so the square matrix \(P\) is invertible.
4. One has\[ H=P^{-1}P^{-T}, \]where\[ P^{-T}\coloneqq(P^{-1})^T. \]
Multiply the identity \[ PHP^T=I \] from step 2 on the left by \(P^{-1}\) and on the right by \((P^T)^{-1}=P^{-T}\). This gives \[ H=P^{-1}P^{-T}. \]
5. Therefore\[ \boxed{H^{-1}=P^TP}. \]
The right-hand side in step 4 is a product of invertible matrices, so \(H\) is invertible. Inverting that identity and reversing the order of factors gives \[ \begin{aligned} H^{-1} &= (P^{-1}P^{-T})^{-1}\\ &= (P^{-T})^{-1}(P^{-1})^{-1}\\ &= P^TP. \end{aligned} \]
6. Q.E.D.
Step 5 is the required identity.