# If $T$ is a self-adjoint operator then $(T-\alpha I)^2+\beta^2I$ is invertible if $\alpha, \beta \in \mathbb{R}$, $\beta\neq 0$

I am trying to show that if $T$ is a self-adjoint operator on a finite-dimensional inner product space $V$ then $(T-\alpha I)^2+\beta^2I$ is invertible if $\alpha, \beta \in \mathbb{R}$, $\beta\neq 0$.

There are several ways I can show $(T-\alpha I)^2+\beta^2I$ is invertible...I can show that $\text{null}(T-\alpha I)^2+\beta^2I=\{0\}$, that $(T-\alpha I)^2+\beta^2I$ has no zero eigenvalues, or that the operator is surjective.

There are a few pieces of information that I have extracted:

a) Since $T$ is self-adjoint, so is $T-\alpha I$ since $(T-\alpha I)^*=T^*-\alpha I^*=T-\alpha I$, since $a\in \mathbb{R}$.

b) I also know that self-adjoint operators only have real eigenvalues.

c) $-\beta^2$ is an eigenvalue of $(T-\alpha I)^2$.

Based on what I know, I still cannot solve the problem. Could someone point me in the right direction? Thank you!

• My attempt: Let $\{\lambda_i\}$ be the set of eigenvalues of $T$. Then $\{\lambda_i-\alpha\}$ is the set of eigenvalues of $T-\alpha I$ and you can show that $\{(\lambda_i-\alpha)^2\}$ are the eigenvalues of $(T-\alpha I)^2$. – polfosol Feb 1 '17 at 12:58
• So the set of eigenvalues of $(T-\alpha I)^2+\beta^2 I$ is $\{(\lambda_i-\alpha)^2+\beta^2\}$ – polfosol Feb 1 '17 at 13:06

1. OP's operator $A$ can be written as $$A~:=~S^{\ast}S+\beta^2 I, \qquad S~:=~T-\alpha I.$$
2. Then $A$ is positive definite: $$\forall x\in V\backslash\{0\}:~~ \langle x | Ax \rangle~=~\underbrace{|| Sx||^2}_{\geq 0} + \underbrace{\beta^2}_{>0} \underbrace{||x||^2}_{> 0 } ~>~ 0 .$$
3. Therefore $${\rm ker}(A)~=~\{0\}.$$
4. Since $V$ is finite-dimensional, then $A$ is invertible.
• $\uparrow$ Yes. – Qmechanic Feb 2 '17 at 14:09