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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)).

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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 $Tv_2=2v_2$...

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  • $\begingroup$ For II, it is true if T is normal correct? $\endgroup$
    – TheNotMe
    Jul 14 '13 at 19:22
  • $\begingroup$ Yes it's correct: $T$ is normal iff $T$ s diagonalizable in an orthogonal basis. $\endgroup$
    – user63181
    Jul 14 '13 at 19:39
  • $\begingroup$ Yes yes I am sorry, deleted my comment haha. Thank you! $\endgroup$
    – TheNotMe
    Jul 15 '13 at 13:34
  • $\begingroup$ You're welcome. $\endgroup$
    – user63181
    Jul 15 '13 at 13:34
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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.

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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.

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  • $\begingroup$ Thanks. That reminded me on how to typeset the duality braces. I was groping a bit and used inequality symbols! $\endgroup$
    – Karthik C
    Jul 14 '13 at 15:57
  • $\begingroup$ Happy to help! ${}$ $\endgroup$ Jul 14 '13 at 15:57

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