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Let $H$ be a Hilbert space (with the inner product $\langle \cdot,\cdot\rangle $), let $x_1,...,x_n$ be a finite sequence of different members of $H$ and let $M:=span\{x_1,...,x_n\}$. It is easy to show that $$T:M\rightarrow M, T(x):=\sum_{k=1}^n \langle x,x_k\rangle x_k, (x\in M)$$ is a topological isomorphism, where to show that $T$ and the inverse $T^{-1}$ are bounded we do not need an explicit formula for $T$ and $T^{-1}$, since $dim \ M\leq n$.

However, I wonder if one can find an explicit formula for $T^{-1}$, namely, given any $y\in M$, is it possible to solve the equation $\displaystyle\sum_{k=1}^n \langle x,x_k\rangle x_k=y$ for $x$ in order to express $x=T^{-1}y$ explicitly (in terms of, possibly, $y$ and $x_k$)?

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    $\begingroup$ Since you are choosing a finite sequence, is this not equivalent to just finding an explicit formula for the inverse of a matrix? $\endgroup$
    – whpowell96
    Commented Jun 9 at 18:05
  • $\begingroup$ @whpowell96, I think, yes. $\endgroup$
    – serenus
    Commented Jun 9 at 18:06
  • $\begingroup$ Well then, once you have an explicit formula for your matrix use Cramer's Rule. $\endgroup$ Commented Jun 9 at 18:28
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    $\begingroup$ The operator is of a particular form from which its positive definitnesss follows. $\endgroup$ Commented Jun 9 at 18:44
  • $\begingroup$ Perhaps it is worthwhile to look for the inverse operator of the same form $\sum_{k=1}^n\langle x,y_k\rangle y_k,$ I would first try the case when $x_k$ are linearly indpendent. $\endgroup$ Commented Jun 11 at 6:07

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