Decide if the following statement is true or false by briefly justify the answer.
Let $(V, \phi)$ be a real Euclidean space of dimension $n$, and let $f: V \to V$ be an orthogonal operator. Then the operator $f-3I$ is an isomorphism (where $I$ = identity).
Could you give me any hint?
Suppose that $f - 3I$ is no isomorphism. Then the equation $fv = 3v$ has a non-trivial solution $v \neq 0$. This is equivalent to the fact that $f$ has eigenvalue $3$. But since $f$ is orthogonal, all eigenvalues of $f$ lie on the unit circle, i.e., they have absolute value $1$. Thus, $3$ can't be an eigenvalue of $f$.
The idea of the proof that all eigenvalues $\lambda$ of $f$ orthogonal fulfill $\lvert \lambda \rvert = 1$ is discussed e.g. in this question.