Prove that V is an identity operator Recently I have attended linear algebra course, unfortunately it was more focused on practice, when I've discovered that I decided to focus more on theory to boost up my skills during summer vacation. Yesterday I've came across following task and literally have no idea how to solve it. Can you help me with this one, or set some direction to digest over?
It is known that linear operator V which is applied in unitary space obtains unitary matrix $V_e$ in basis $e$ and positive-definite matrix $V_f$ in basis $f$. The task is to prove that $V$ is identity operator.
 A: Fix any basis $b$ and consider the matrix $A=V_b$. Then you know that there exist invertible matrices $S$ and $T$ such that $S^{-1}AS$ is unitary and $T^{-1}AT$ is positive definite.
Since similar matrices share eigenvalues, you see that the eigenvalues of $A$ have modulus $1$ (because $S^{-1}AS$ is unitary) and positive real (because $T^{-1}AT$ is positive definite).
This only leaves tha possibility that $A$ has the single eigenvalue $1$.
Now use that positive definite matrices are diagonalizable.
A: We have that $V$ is unitary and that $V$ is self-adjoint and positive definite. Hence
$$V^{-1}= V^*$$
and
$$V=V^*.$$
This gives $V^2=I$ thus
$$(V-I)(V+I)=0.$$
Since $V$ is positive definite, $-1$ is not an eigenvalue of $V$, therefore $V+I$ is invertible. This gives
$$V=I.$$
A: $V_f$ has the same spectrum as $V_e$ because both matrices are matrix representations of the same linear operator $V$ in some bases. Hence the eigenvalues of $V_f$ are positive real numbers (because $V_f$ is positive definite) with unit moduli (because $V_e$ is unitary), i.e. all eigenvalues of $V_f$ are equal to $1$. Hence $V_f=I$ and $V$ is the identity operator.
