Prove/disprove the following theories regarding operators in inner product spaces

Question

Prove/disprove:

(I) Let $V$ be a vector space with an inner product upon field $F$. Given an operator $T:V\to V$, which is invertible. Is $T^{*}$ invertible?

(II) Let $v_1, ..., v_k$ be eigenvectors of $T$ that are correspondent to different eigenvalues, is $\{v_1,...,v_k\}$ and orthogonal set? (In the same vector space from (I)).

Answer 1

1) Yes if $T$ is invertible with inverse $T^{-1}$ then $T^*$ is invertible with inverse $(T^{-1})^*$: $$TT^{-1}=\mathrm{id}\Rightarrow (T^{-1})^*T^*=\mathrm{id}$$

2) It's easy to construct a counterexample. For example in $\mathbb R^2$ take any two linearly independant vectors $v_1,v_2$ and define $T$ by $T v_1=v_1$ and $Tv_2=2v_2$...

Answer 2

First:

By definition, $\langle Tu,v\rangle=\langle u,T^*v\rangle$ If $T^*v=T^*w$, then by above, $\langle Tu,v \rangle=\langle Tu,w\rangle$ or equivalently, $\langle Tu,v-w\rangle=0$ must hold for all $u$. So by invertibility of $T$ and non-degeneracy, $v=w$ and hence $T^*$ is injective, whence by dimension count, $T^*$ is an isomorphism.

Second:

It's not necessary that eigenvectors corresponding to different eigenvalues form an orthogonal set. One can construct counter-examples.

Answer 3

For (I): \begin{align*} w \in \ker T^* &\iff T^* w = 0 \\ &\iff \langle v, T^* w\rangle = 0 \quad \forall v \in T \\ &\iff \langle Tv, w\rangle = 0 \quad \forall v \in T \\ &\iff w \in \left(\operatorname{im}T\right)^\perp \end{align*}

It follows that: $$ \dim \ker T^* = \dim V - \dim \operatorname{im} T $$

If $T$ is invertible, then $\dim V = \dim \operatorname{im} T$. Thus, $\dim \ker T^* = 0$. It follows that $T^*$ is injective, and hence invertible.

For (II):

Consider $V = \mathbb R^2$, $T(x, y) = (x + y, 2y)$. We have $T(1, 0) = (1, 0)$ and $T(1, 1) = 2(1, 1)$. Yet, $(1, 0)$ and $(1, 1)$ are not orthogonal.

This claim is true, however, if $T$ is normal.