# Eigenvalues of the Adjoint of a Linear Map

In the following proposition we will see that the eigenvalues of $T^*$ are the complex conjugate eigenvalues of $T$.

Proposition 1: Let $V$ be a finite-dimensional nonzero inner product spaces. Then $\lambda$ is an eigenvalue of $T$ if and only if $\overline{\lambda}$ is an eigenvalue of $T^*$. |

**Proof:**We will prove the contrapositives to proposition 1.

- $\Rightarrow$ First suppose that $\lambda$ is not an eigenvalue of $T$. Then $(T – \lambda I)$ is injective which implies that $(T – \lambda I)$ is invertible, and so there exists a linear map $S \in \mathcal L (V)$ such that $S(T – \lambda I) = I = (T – \lambda)S$. If we take the adjoint of both sides of the equation above, then we have that:

(1)

\begin{align} \quad S(T – \lambda I) = I = (T – \lambda)S \\ \quad (S(T – \lambda I))^* = I^* = ((T – \lambda)S)^* \\ \quad S^*(T – \lambda I)^* = I = (T – \lambda)^* S^* \end{align}

- Thus $(T – \lambda I)^*$ is invertible which implies that $(T – \lambda I)^* = (T^* – \overline{\lambda}I)$ is injective, so $\overline{\lambda}$ is not an eigenvalue of $T^*$.

- $\Leftarrow$ Now suppose that $\overline{\lambda}$ is not an eigenvalue of $T^*$. Then $(T^* – \overline{\lambda}I) = (T – \lambda I)^*$ is injective which implies that $(T – \lambda I)^*$ is invertible, and so there exists a linear map $S^* \in \mathcal L (V)$ such that $S^* (T – \lambda I)^* = I = (T – \lambda I)^* S^*$. If we take the adjoint of both sides of the equation above, then we have that:

(2)

\begin{align} \quad S^* (T – \lambda I)^* = I = (T – \lambda I)^* S^* \\ \quad (S^* (T – \lambda I)^*)^* = I^* = ((T – \lambda I)^* S^*)^* \\ \quad S^{**} (T – \lambda I)^{**} = I = (T – \lambda I)^{**} S^{**} \\ \quad S ( T – \lambda I) = I = (T – \lambda I)S \end{align}

- Therefore $(T – \lambda I)$ is invertible which implies that $(T – \lambda I)$ is injective and so $\lambda$ is not an eigenvalue of $T$. $\blacksquare$

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