# Need a help in finding the inverse of an operator .

The question and part of its answer is given as follows:

13. Let $K$ be an operator of a finite rank on a Hilbert space $H$. For $\varphi \in H$, $$K\varphi = \sum_{j=1}^{n} \langle \varphi, \varphi_j\rangle\psi_j.$$ Suppose $\psi_j \in \operatorname{sp}\{ \varphi_1, \cdots, \varphi_n\}^{\perp}$ for $j = 1, \cdots, n$. Prove that $\mathrm{I} + \alpha K$ is invertible for any $\alpha$ and find its inverse.

Solution. Let $\alpha \in \mathbb{C}$, $K'\varphi = \sum_{j=1}^{n} \langle \varphi, \varphi_j\rangle (-\alpha \psi_j)$. By Theorem 7.1, \begin{align*} \text{$\mathrm{I} + \alpha K$ is invertible} &\quad \Leftrightarrow \quad \text{$\mathrm{I}-K'$ is invertible} \\ &\quad \Leftrightarrow \quad \det(\delta_{ij}-\langle(-\alpha\psi_j),\varphi_i\rangle)_{i,j=1}^{n}\neq 0. \end{align*} But $$\det(\delta_{ij}-\langle(-\alpha\psi_j),\varphi_i\rangle)_{i,j=1}^{n} = \det(\delta_{ij}+\alpha\langle\psi_j,\varphi_i\rangle)_{i,j=1}^{n} = 1 \neq 0$$ (because $\psi_j \in \operatorname{sp}\{ \varphi_1, \cdots, \varphi_n\}^{\perp}$ for $j = 1, \cdots, n$). Thus $\mathrm{I}+\alpha K$ is invertible.

A part of the theorem is given as follows:

7.1. Theorem. Suppose $K \in L(H)$ is of finite rank, say $$Kx = \sum_{j=1}^{n} \langle x, \varphi_j\rangle\psi_j.$$ The operator $I-K$ is invertible if and only if $$\det(\delta_{ij}-\langle\psi_j,\varphi_i\rangle)_{i,j=1}^{n} \neq 0.$$ In this case, for every $y \in H$, $$(I-K)^{-1}y = y - \frac{1}{\det(a_{ij})}\det\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & \langle y, \varphi_1\rangle \\ a_{21} & a_{22} & \cdots & a_{2n} & \langle y, \varphi_2\rangle \\ \vdots & \vdots & & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} & \langle y, \varphi_n\rangle \\ \psi_1 & \psi_2 & \cdots & \psi_n & 0 \end{pmatrix}$$

Then the action of the inverse of the operator on $y$ is $y-I$, but how can this tell me what is the inverse of the operator $K$, could anyone clarify this for me please?

Thanks!

Your hypotheses on $\{\psi_1,\dotsc,\psi_n\}$ and $\{\varphi_1,\dotsc,\phi_n\}$ make an appeal to that theorem completely and utterly unnecessary. Since $\{\psi_1,\dotsc,\psi_n\} \subset \{\varphi_1,\dotsc,\phi_n\}^\perp$, it immediately follows that $K^2 = 0$. What, then, is $I^2 - \alpha^2 K^2$, and why should this suggest to you an explicit inverse for $I + \alpha K$?

• you mean that it is not important for me to use this theorem at all?
– user426277
Commented Dec 26, 2017 at 16:57
• Your second statement is not clear for me at all (the statement containing $K^2 = 0$) could you explain it in details please?
– user426277
Commented Dec 26, 2017 at 17:00
• Frankly speaking I do not an answer to your last question ( why should this suggest to you an explicit inverse for ....?) :( ... could u please explain the answer for me ?
– user426277
Commented Dec 26, 2017 at 17:02
• (1) You can explicitly check that $K^2 \phi = 0$ for all $\phi \in H$ using the hypothesis that $\{\psi_1,\dotsc,\psi_n\} \subset \{\varphi_1,\dotsc,\varphi_n\}^\perp$. (2) How would you usually factorise $x^2-y^2$? Commented Dec 26, 2017 at 17:51
• you mean that the inverse is $(I - \alpha K)$?
– user426277
Commented Dec 26, 2017 at 19:13