# How to find an explicit formula for an inverse of a linear operator

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 $$x_k$$)?

• Since you are choosing a finite sequence, is this not equivalent to just finding an explicit formula for the inverse of a matrix? Commented Jun 9 at 18:05
• @whpowell96, I think, yes. Commented Jun 9 at 18:06
• Well then, once you have an explicit formula for your matrix use Cramer's Rule. Commented Jun 9 at 18:28
• The operator is of a particular form from which its positive definitnesss follows. Commented Jun 9 at 18:44
• 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. Commented Jun 11 at 6:07