Answer to Question #211220 in Functional Analysis for Muhammad

Question

 Let X be a finite dimensional inner product space and T : X → X be

a linear operator. If T is self adjoint (that is < x, T x >=< T ∗x, x >).

Show that its spectrum is real. If T is unitary then show that its

eigenvalues have absolute value 1.

Answer 1

A linear operator T is self adjoint if and only if for all x $\langle Tx, x \rangle = \langle x, Tx \rangle$ .

Let $\lambda$ be any eigenvalue of T and x a corresponding eigenvector. Then

$\langle Tx, x \rangle = \langle \lambda x, x \rangle = \bar\lambda |x|^2$

$\langle x, Tx \rangle = \langle x, \lambda x \rangle = \lambda |x|^2$

Compare these expressions, reminding that $\langle Tx, x \rangle = \langle x, Tx \rangle$. We obtain

$(\bar\lambda-\lambda) |x|^2=0$ , and hence $\bar\lambda=\lambda$. This means that eigenvalues of T are real.

A linear operator T is unitary if and only if it is invertible and for all x $\langle T^{-1}x, x \rangle = \langle x, Tx \rangle$.

Let $\lambda$ be any eigenvalue of T and x a corresponding eigenvector. Then x is also an eigenvector of $T^{-1}$ with the eigenvalue $\lambda^{-1}$. Compute

$\langle T^{-1}x, x \rangle = \langle \lambda^{-1} x, x \rangle = \bar\lambda^{-1} |x|^2$

$\langle x, Tx \rangle = \langle x, \lambda x \rangle = \lambda |x|^2$

Compare these expressions, reminding that $\langle T^{-1}x, x \rangle = \langle x, Tx \rangle$. We obtain

$(\bar\lambda^{-1}-\lambda) |x|^2=0$ , and hence $\bar\lambda^{-1}=\lambda$. This means that $|\lambda|^2=\bar\lambda\lambda=1$.