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This question already has an answer here:

Let $X$ be an inner product space such that $\dim X < \infty$, and let $T \colon X \to X$ be an isometric linear operator.

Since $\dim X < \infty$, $X$ is complete and thus a Hilbert space; since $T$ is isometric, $T$ is also injective and hence also surjective and thus bijective, because $\dim X < \infty$. So $T^{-1}$ exists.

How to show that $T$ is unitary. That is, how to show that the Hilbert adjoint operator $T^*$ of $T$ equals $T^{-1}$?

Since $X$ is finite-dimensional, we can choose an orthonormal basis for $X$; let $n \colon= \dim X$, and let $\{e_1, \ldots, e_n \}$ be an orthonormal basis for $X$.

Then, for each $i, j = 1, \ldots, n$, we have $$\langle Te_i , e_j \rangle = \langle e_i, T^* e_j \rangle,$$ and, $$\langle T^* T e_i, e_j \rangle = \langle T e_i , T e_j \rangle = \langle e_i , e_j \rangle = \begin{cases} 1 \ & \mbox{ if } \ i = j \\ 0 \ & \mbox{ if } \ i \neq j. \end{cases} $$

What next?

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marked as duplicate by DisintegratingByParts, Venus, user147263, Lord_Farin, kjetil b halvorsen Jun 6 '15 at 19:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @T.A.E., yes, you're right? Thanks. $\endgroup$ – Saaqib Mahmood Jun 6 '15 at 12:14
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This is almost a repost, see Linear surjective isometry then unitary

The calculation the poster makes is your missing link :)

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