Suppose $T \in L(V)$ and $\lambda \in \mathbb{F}$. Prove that $\lambda$ is an eigenvalue of $T$ iff $\overline{\lambda}$ is an eigenvalue of $T^\ast$.

Question

I would like to make sure I'm right with the proof below.

Any corrections or suggestions are welcomed!!

My proof:

Suppose $\lambda$ is an eigenvalue of $T$.

$\Rightarrow \exists v\neq0$ s.t $Tv=\lambda v$

Now, consider $<Tv,w>$ for all $w \in V$.

Note that $<Tv,w> = <v,T^\ast w>=<\lambda v,w>=<v,\overline{\lambda}w> \forall w \in V.$

$\Rightarrow \left \langle v,T^\ast w - \overline{\lambda}w \right \rangle=0$ for all $w \in V$

$\Rightarrow \left \langle v, (T^\ast - \overline{\lambda} I)w \right \rangle=0$ for all $w \in V$

$\Rightarrow T^\ast - \overline{\lambda} I =0$

$\Rightarrow T^\ast v = \overline{\lambda} v$

Since $v \neq 0$, $\overline{\lambda}$ is an eigenvalue of $T^\ast$ as desired.

Conversely, suppose $\overline{\lambda}$ is an eigenvalue of $T^\ast$ with an eigenvector $v$.

Now, consider $<w, T^\ast v>$ for all $w \in V$.

Note that $<w,T^\ast v>=<Tw, v>=<w,\overline{\lambda}v>=<\lambda w, v>$

$\Rightarrow \left \langle Tw - \lambda w,v \right \rangle =0$ for all $w \in V$

$\Rightarrow$ By similar reason above, $T-\lambda I=0$

$\Rightarrow Tv=\lambda v$

Since $v \neq 0$, $\lambda$ is an eigenvalue of $T$.

Hence, $\lambda$ is an eigenvalue of $T$ iff $\overline{\lambda}$ is an eigenvalue of $T^\ast$.

Answer 1

Your proof doesn't work. There are two major problems.

1. $\langle v , Aw \rangle = 0$ for all $w\in V$ does not imply that $A = 0$. The reason is simple, you have a fixed vector $v$, all this tells you that the range of $A$ is orthogonal to the subspace generated by $v$. Since, in general, $v$ doesn't span the whole $V$, you can't conclude that the range of $A$ is $\{0\}$.
2. Even if the above did work (which it doesn't), you do not want to prove that $T^*-\overline\lambda I = 0$. You want to prove that there exists non-zero vector $w$ such that $(T^*-\overline\lambda I)w = 0$, which is entirely different thing (universal vs existential quantification).

So, in 2. I brought up that you want to prove that $(\exists w\in V\setminus\{0\})\ (T^*-\overline\lambda I)w = 0.$ In other words, $\ker(T^*-\overline\lambda I) \neq \{0\}$. By rank and nullity theorem, this is the same as proving that $T^*-\overline\lambda I$ is not surjective. Assume the contrary, that it is surjective. Then, as you showed, we would have $\langle v, (T^*-\overline\lambda I)w \rangle = 0$, for all $w\in V$, but since we assumed that $T^*-\overline\lambda I$ is surjective, it means that $v$ is orthogonal to the whole $V$. But this is only possible if $v = 0$, which is contradiction with the assumption that $v$ is an eigenvector, and therefore non-zero.

Now we have $\lambda$ eigenvalue of $T$ implies that $\overline\lambda$ is eigenvaule of $T^*$. The other direction follows immediately by applying the proven implication on $T^*$ instead of $T$ and noting that $(T^*)^* = T$ and $\overline{\overline{\lambda}} = \lambda$.

Answer 2

Here is a direct proof that does not rely on contradiction.

$\rightarrow$ Suppose that $\lambda$ is an eigenvalue of the linear map $T\in \mathscr{L}(V)$. Then, there exists a nonzero vector $v\in V$ such that $Tv = \lambda v$.

Moreover, there exists a vector $w\in V$ and a linear map $T^{*}\in \mathscr{L}(V)$ such that $<Tv,w> = <v,T^{*}w>$. (note that $T^{*}w$ is unique, so you cannot say anything about "all" $w\in V$)

Then, we use properties of inner products to do some algebra:

$$\begin{aligned} <Tv,w> &= <v,T^{*}w> \\ &= <\lambda v, w> (\text{definition of eigenvalue}) \\ &= \lambda <v,w> (\text{linearity in first slot}) \\ &= \lambda \overline{<w,v>} (\text{conjugate symmetry}) \\ &= \overline{\overline{\lambda} <w,v>} (\text{definition of conjugate}) \\ &= \overline{<\overline{\lambda} w,v>} (\text{linearity in first slot}) \\ &= <v,\overline{\lambda} w> (\text{conjugate symmetry}) \end{aligned}$$

So, equating the first line with the last line, we have

$$ <v,T^{*}w> = <v,\overline{\lambda} w> $$

Thus, we have found a vector $w\in V$, such that $T^{*}w = \overline{\lambda} w$. This implies that $\overline{\lambda}$ is an eigenvalue for $T^{*}$.

$\leftarrow$ We can use the last paragraph of Ennar's proof to conclude.

Now we have $\lambda$ eigenvalue of $T$ implies that $\lambda$ is eigenvaule of $T^{*}$. The other direction follows immediately by applying the proven implication on $T^{*}$ instead of $T$ and noting that $(T^{*})^{*}=T$ and $\overline{\overline{\lambda}}$=$\lambda$.