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.


2 Answers 2


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, 2015 at 7:38
  • $\begingroup$ Which part of this argument required finite dimensionality? $\endgroup$ Jul 29 at 17:12
  • $\begingroup$ @BallpointBen If $V$ is not finite-dimensional, then $T^*$ is not necessarily defined everywhere on $V$. Furthermore, if $V$ is not finite-dimensional, then $\lambda$ is an eigenvalue of $T$ is not equivalent to $T - \lambda I$ is not invertible. $\endgroup$ Jul 29 at 21:50
  • $\begingroup$ @SheldonAxler $\lambda $ being an eigenvalue of a linear map $T$ implies that there exists a non-zero vector $v$ such that $(T-\lambda I)v=0$. Doesn’t this imply that $T-\lambda I$ is not Injective and therefore not invertible as a linear map is invertible if and only if it is injective and surjective? I don’t see the need for it to be finite dimensional. I’m pretty sure I’m missing something obvious but can’t see what. $\endgroup$
    – Seeker
    Aug 31 at 22:12
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    $\begingroup$ @Seeker You are correct that if $\lambda$ is an eigenvalue of $T$, then $T - \lambda I$ is not invertible. This then implies that $T^* - \bar{\lambda} I$ is not invertible, as shown above. However, you then need to assume that $V$ is finite-dimensional to reach the desired conclusion that $\bar{\lambda}$ is an eigenvalue of $T^*$. $\endgroup$ Sep 1 at 2:07

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.

  • $\begingroup$ Why do we need to put a restriction that $|\lambda| < 1$? For any $\lambda$, can we not take eigenvector $v = (a_1, \lambda a_1, \cdots)$ just as you did in your proof? $\endgroup$
    – James C
    Aug 29 at 15:14
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    $\begingroup$ No, we cannot, because $v$ needs to be an element of the Hilbert space $l_2$, and in particular, $\sum_{n=1}^{\infty}\lambda^{2n}$ must converge. $\endgroup$ Aug 30 at 8:50

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