Proof of: If $T$ is self adjoint, then $\operatorname{id}_V+T^2$ is an invertible linear transformation

Question

So I want to study the spectral theorem and self-adjoint matrices.

I am new to the topic, and I am trying to solve problems dealing with them. One of these problems is this one:

Help, please?

Answer 1

By the spectral theorem, the matrix of $T$ with respect to some basis is diagonal and the entries of the main diagonal are real numbers. Therefore the matrix of $\operatorname{id}_V+T^2$ with respect to the same basis is diagonal and the entries of the main diagonal are all greater than or equal to $1$. Therefore, its determinant is greater than $0$ (an, in particular, different from $0$).

However, if you take $T\colon\mathbb{R}^2\longrightarrow\mathbb{R}^2$ define by $T(x,y)=(-y,x)$, then $\operatorname{id}_V+T^2$ is the null function.

Answer 2

a) without the spectral theorem: let $ \mu$ be an eigenvalue of $\operatorname{id}_V+T^2$, then there is an eigenvalue $ \lambda$ of $T$ such that $ \mu=1+\lambda^2$. Since $T$ is selfadjoint, $ \lambda $ is real, hence $\mu \ge 1 >0$.

Therefore $\operatorname{id}_V+T^2$ is invertible.