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I am studying Image Processing in which unitary transforms play an important role, one reason that I found for their use in image transformation is isometry(they preserve distance), I found a relevant answer for this on the same site The difference between an isometric operator and a unitary operator on a Hilbert space?. Where to prove the same author starts with, $$||Tx-Ty||=||x-y||$$this can be proven by $$|T|||x-y||$$ where $|T|=1$ and T is unitary, but here he further adds on to prove that

"Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $$⟨T^HTx,x⟩=⟨x,x⟩\quad\quad \forall x$$ This implies that $T^H T=I$. Conversely, if $T^H T=I$, you can show that T is an isometry"

Now this proof is necessary other wise any matrix with determinant 1 can be used for transformation, which off course isn't true. Can somebody help me with the missing link, I am not an expert in linear algebra and just know basics of norm.

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Userhanu
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  • If you are familiar with the proof in real Euclidean spaces that orthogonal transformations preserve length (isometry), the proof for unitary transformations is very similar. – hardmath May 12 '16 at 15:22
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    I'm not sure I understand exactly what your question is. The answer at the link does not claim that $|T(x-y)| = \text{det}(T)|x-y|$ as you seem to be asserting. (In fact, this is surely false in general, as can be seen by taking any noninvertible $T$ where $x-y$ is not in the null space of $T$.) –  May 12 '16 at 15:26
  • That is what I want to say i am somewhere wrong in proving, because if that does happen then any matrix with 1 as a determinant can be used for image transformations which is not true. – Userhanu May 12 '16 at 15:29
  • Well, if your proof relies on the claim that $|T(x-y)| = \text{det}(T)|x-y|$, then this is wrong even if $\text{det}(T) = 1$. For example, take $T = \begin{pmatrix}1 & 1 \ 0 & 1 \ \end{pmatrix}$, $x = \begin{pmatrix}1 \ 1 \end{pmatrix}$, and $y = \begin{pmatrix}0 \ 0 \end{pmatrix}$. –  May 12 '16 at 15:34
  • $T^T$ isn't referring to $TT$ it means $T^H\times T$ or $T^* \times T$ – Patrick Abraham May 12 '16 at 15:43
  • I think I can prove it by $$(Ux-Uy)^{T}(Ux-Uy)=(x-y)^{T}(x-y)$$ if that is correct – Userhanu May 12 '16 at 15:45

1 Answers1

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First, notice that if we have two vectors $$ v = \begin{pmatrix} v_1 \\ v_2 \\v_3 \end{pmatrix} \ \ w = \begin{pmatrix} w_1 \\ w_2 \\w_3 \end{pmatrix} $$

we can rewrite the dot product $$\langle v, w\rangle = v^H \times w = \begin {pmatrix} v_1^* & v_2^* & v_3^*\end{pmatrix} \times \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix}$$

Now, we wish to show that the operators that preserve distances are exactly the unitary operators.

Let $T: V \to V$ be any operator.

Now, to show that $T$ preserves distances, we need that $$ \langle T v, T w \rangle = \langle v, w \rangle $$

this is because, if $T$ preserves the inner product structure, then it also preserves the metric generated by the inner product ($|v| = \sqrt{\langle v, v \rangle})$

Now,

$$ \langle T v, T w \rangle = (T v)^H \times (T w) = (v^H T^H) (Tw) = v^H (T^HT) w $$

However, we want $$ \langle T v, T w \rangle = \langle v, w \rangle $$

So this implies that,

$$ v^H (T^HT) w = v^H w \\ \implies T^H T = I $$

which tells us that any operator $T$ that preserves the inner product structure (and hence the metric) must be unitary.

Siddharth Bhat
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