16 - Non-Singularity of Matrix of Normal Form

Non-Singularity of Matrix of Normal Form

Coefficient Matrix $A$ from Normal Form $A\alpha =\psi$ is non singular

Proof

Suppose there $\exists \alpha \ne 0$ with $A\alpha =0$ hence

$${\alpha }^{T}A\alpha =0⟺\sum _{i=0}^n{\alpha }_i(A\alpha {)}_i=0⟺\sum _{i=0}^n{\alpha }_i\sum _{k=0}^n{a}_{ik}{\alpha }_k=0$$

As $a_{ik}={\int }_a^{b}w(x)x^k{x}^{i}dx$

$$⟺\sum _{i=0}^n{\alpha }_i\sum _{k=0}^n\left({\int }_a^{b}w(x)x^k{x}^{i}dx\right){\alpha }_k=0$$

Rearranging then

$${\int }_a^{b}w(x)\left(\sum _{i=0}^n{\alpha }_i{x}^i\right)\left(\sum _{k=0}^n{\alpha }_k{x}^k\right)dx=0$$

Thus

$${\int }_a^{b}w(x){\left(\sum _{i=0}^n{\alpha }_i{x}^i\right)}^{2}dx=0$$

Hence

$$\sum _{i=0}^n{\alpha }_i{x}^i=0$$

So

$${\alpha }_i=0\text{ for }i=0,1,\cdots ,n$$

This contradicts original assumption so $A$ is non-singular