I am asked to prove that if $T: V \rightarrow V$ is a linear operator over a complex inner product space $(V,\langle,\rangle)$, then $\overline{\lambda}$ is an eigenvalue of $T^*$ where $\lambda$ is an eigenvalue of $T$.

The problem is easily reduced to the cases where either $V$ is a finite dimensional space or $T$ is a normal operator, but I am stuck with the general case.


As nicely pointed out by uniquesolution, we must assume that $V$ is finite dimensional. However, there is no need to use determinants:

Suppose $T$ is a linear operator on a finite-dimensional complex vector space $V$ and $\lambda \in \mathbf{C}$. Note that $\lambda$ is not an eigenvalue of $T$ if and only if $T - \lambda I$ in invertible, which happens if and only if there exists an operator $S$ on $V$ such that $$ S(T - \lambda I) = (T - \lambda I)S = I. $$ Taking adjoints of all three sides above shows that the equations above are equivalent to $$ (T^* - \bar{\lambda} I)S^* = S^*(T^* - \bar{\lambda} I) = I. $$ Thus we see that $T - \lambda I$ is invertible if and only if $T^* - \bar{\lambda} I$ is invertible. In other words, $\lambda$ is an eigenvalue of $T$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.

  • $\begingroup$ nice argument ! $\endgroup$ Sep 12 '15 at 7:38

The result is true if $V$ is finite-dimensional, because in that case, $\lambda\in\mathbb{C}$ is an eigenvalue of $T$ if and only if $\det(T-\lambda I)=0$, the determinant of the adjoint is the complex conjugate determinant:

$$\det(T-\lambda I)=\overline{\det(T^*-\overline{\lambda}I)}$$ so $\det(T^*-\overline{\lambda}I)=0$, whence $\overline{\lambda}$ is an eigenvalue of $T^*$.

If $V$ is infinite dimensional, the result no longer holds. Take for example $V=l_2$, an infinite dimensional Hilbert space over the complex numbers, and let $$T(a_1,a_2,a_3,\dots )=(a_2,a_3,\dots)$$ $T$ is the left shift, and every $\lambda$ such that $|\lambda| <1$ is an eigenvalue of $T$ (just pick any $a_1\neq 0$ and take $v=(a_1,\lambda a_1,\lambda^2 a_1,\dots)$). The adjoint of the left shift is the right shift


which has no eigenvalues at all.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.