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We all know that a Gram matrix (a matrix with entries that are inner products of basis functions) is a invertible. Suppose I have $a_{ij} = (h_i, h_j)_H$ where the $h_j$ are basis functions of a Hilbert space $H$. Is the infinite matrix $A=(a_{ij})$ invertible? That is, is $\sum_{j=1}^\infty a_{ij}v_i = b_i$ solvable for $v_i$ for each $i$ given $b_i$?

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Yes.

For an arbitrary element $x$, let $x^* \in H^*$ denote $(\cdot,x)_H$.

Let $V$ denote the transformation $$ Vx = \sum_{i} x_i h_i $$ That is, $V$ is the matrix whose "column vectors" are the basis functions $h_i$. Note that $A = V^*V$.

Now, note that for any $x$, $x^*(Ax) = (x^* V^*)(Vx) = (Vx)^*(Vx) = \|Vx\|_H^2$. It follows that $Ax = 0$ if and only if $Vx = 0$. Since $V$ has a basis as its columns, $V$ is indeed invertible.

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  • $\begingroup$ thanks but I am confused about what $V$ is. $V:H \to H$ but how does $x_ih_I$ make sense?? $\endgroup$
    – user173805
    Sep 8 '14 at 21:03
  • $\begingroup$ Select an orthonormal basis $\{v_i\}$. Define $V : H \to H$ by $\sum x_i v_i \mapsto \sum x_i h_i$; that is, by $v_i \mapsto h_i$. $\endgroup$ Sep 8 '14 at 21:07
  • $\begingroup$ ...........thank u $\endgroup$
    – user173805
    Sep 8 '14 at 21:15

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