prove that $\lambda$ is eigenvalue of $T$ if and only if $\bar{\lambda}$ is eigenvalue for $T^*$

Question

Suppose $\lambda\in K$ and $T\in L(V)$ where $V$ is finite dimensional inner-product space over $K$ .Prove that prove that $\lambda$ is eigenvalue of $T$ if and only if $\bar{\lambda}$ is eigenvalue for $T^*$

$<Tv,u> = <\lambda v,u> = \bar\lambda<v,u>=<v,\bar\lambda u>$

how to solve i am not geeing any idea

Answer 1

The proof comes down to showing that if $T$ has an eigenvector, then so does $T^*$:

Suppose that $v$ is an eigenvector of a linear operator $T$ on a finite dimensional inner-product space $K$ with corresponding eigenvalue $\lambda$. Then, for arbitrary vector $x \in K$,

$0=<0,x>=<(T-\lambda I)(v),x)>=<v,(T-\lambda I)^*, (x))> = <v,(T^* -\bar{\lambda} I),(x)>$

From here, notice that $v$ is orthogonal to the range of $(T^* -\bar{\lambda} I)$.

What can you say from here about the kernel?

Answer 2

$$\langle\,T^*u, v\,\rangle = \overline{\langle\,Tv, u\,\rangle} = \overline{\langle\,\lambda v, u\,\rangle} = \overline{\bar{\lambda}\langle\,v, u\,\rangle} = \lambda \langle\,u,v\,\rangle = \langle\, \bar{\lambda}u,v\,\rangle $$ Going the other way: $$\langle\,Tv, u\,\rangle = \overline{\langle\,T^*u, v\,\rangle} = \overline{\langle\,\bar{\lambda} u, v\,\rangle} = \overline{\lambda\langle\,u, v\,\rangle} = \bar{\lambda} \langle\,v,u\,\rangle = \langle\, \lambda v,u \,\rangle $$