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So, I'm looking at this proof, and it makes no sense to me at all.

Theorem: Let $V$ be a finite dimensional vector space over $\mathbb{R}$, with a positive definite scalar product. A linear map $A \colon V \to V$ is unitary if and only if

$$A^tA=I.$$

Proof: The operator $A$ is unitary if and only if

$$\langle Av, Aw \rangle = \langle v, w \rangle$$

for all $v,w \in V$. This condition is equivalent with

$$\langle A^tAv, w \rangle = \langle v, w \rangle.$$

Just stopping right there, I do not get why those two conditions are equivalent.

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How do you define $A^T$? It's the matrix such that

$$\langle x, Ay \rangle = \langle A^Tx, y \rangle, \quad \text{for all $x,y$}.$$

See here.

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  • $\begingroup$ Oh it meant the adjoint. Well... haha. $\endgroup$ – user104436 Nov 13 '13 at 23:58
  • $\begingroup$ I assume so, because otherwise it would be true (it's not the usual transpose in all scalar products). I must admit that using $A^t$ is confusing. $\endgroup$ – Vedran Šego Nov 13 '13 at 23:59
  • $\begingroup$ Well, it is the adjoint because the vector space is a real one...If it were a complex one it'be be $\;A^*=\overline{A^t}\;$ $\endgroup$ – DonAntonio Nov 14 '13 at 0:08
  • $\begingroup$ @DonAntonio Yes, but a different notification would be nice, since it's a positive scalar product (so, not necessarily the Euclidean one). For example, when I work with the indefinite scalar products, I usually write the product of $x$ and $y$ as $[x,y]$ or $[x,y]_J = y^* J x$, where $J$ induces the scalar product. In that case, I usually denote the adjoint of $X$ as $X^{[*]}$ or $X^{[*]_J} = J^{-1} X^* J$, to make it different from the traditional (conjugate) adjoint $X^*$. $\endgroup$ – Vedran Šego Nov 14 '13 at 0:13
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Certainly the steps shown in the question are OK;

$\langle Av, Aw \rangle = \langle v, w \rangle \tag{1}$

implies

$\langle A^TAv, w \rangle = \langle v, w \rangle; \tag{2}$

next, we take things a little further: from (2) we infer that

$\langle A^TAv - v, w \rangle = 0 \tag{3}$

for all $w$, whence

$A^TAv = v, \tag{4}$

that is,

$A^TA = I. \tag{5}$

The engine also runs in reverse: $(5) \Rightarrow (4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1).$ QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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