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Let $V$ be a vector space with the orthonormal basis $Q = \{ \vec{q_1},\ldots, \vec{q_n} \}$ and let $\ell:V\to V$ be an orthogonal map. Prove that the matrix $L$ of of $\ell$ with respect to $Q$ is orthogonal.

Note: By orthogonal map I mean that $\ell$ is linear and satisfies $\left\Vert \ell(\vec{x}) \right\Vert = \left\Vert \vec{x} \right\Vert$ for all $x \in V$. By orthogonal matrix I mean that $L$ has orthonormal columns.

I have that $ L=\left[ \begin{array}{ccc} [\ell(\vec{q_1})]_Q & \cdots & [\ell(\vec{q_n})]_Q \end{array} \right] $

but I don't have any idea how to proceed.

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    If you are in a real vector space do you mean that $l : V \mapsto V$ is a unitary operator?2011-10-29

2 Answers 2

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A proof sketch.

  1. Show that $\langle \ell(\vec x), \ell(\vec {y}) \rangle = \langle \vec x , \vec y \rangle$ holds for all $\vec x, \vec y \in V$. For this step, you may need the following hint: $\langle \vec x, \vec y \rangle = \frac{1}{2} (\| \vec x+\vec y \|^2 - \| \vec x \|^2 - \| \vec y \|^2) .$

  2. Show that if the $i^{\rm th}$ column of $L$ is $c_i \in \mathbb R^n$, then $ \ell(\vec {q_i}) = \sum_{k = 1}^n c_{i,k} \cdot \vec {q_k}. $

  3. Show that for $1 \leq i, j \leq n$, we have $ \langle \ell(\vec {q_i}) , \ell(\vec {q_j}) \rangle = \langle c_i, c_j \rangle. $

  4. Using (1.), what can you say about $\langle \ell(\vec {q_i}) , \ell(\vec {q_j}) \rangle$? What does this mean about $\langle c_i, c_j \rangle$?

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    @Kb100 Sorry, was out. Do you want to chat? (Please do not hurry to accept an answer. It is usually better to wait out a few days before accepting any answer.)2011-10-29
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Suppose $T: V \to V$ is an isometry, i.e. $\|Tx\| = \|x\|$ for all $x \in V$. Then $T$ is clearly injective, so since $V$ is finite-dimensional, it is also surjective. This means $T^{-1}$ exists. By the polarization identity we have $\langle Tx, Ty \rangle = \frac{1}{2}(\|T(x+y) \|^2- \|Tx\|^2 - \|Ty\|^2)$ $ = \frac{1}{2}(\|x+y \|^2- \|x\|^2 - \|y\|^2) = \langle x, y \rangle$ which implies $\langle Tx, y \rangle = \langle x, T^{-1}y\rangle$ Hence $T^{-1} = T^*$. But the matrix for $T^*$ with respect to an orthornormal basis $Q = \{q_1, ..., q_n\}$ is Hermitian transpose of the matrix for $T$: $[T]_Q^* = [T^*]_Q = [T^{-1}]_Q = [T]_Q^{-1}$ So $[T]_Q[T^*]_Q = [T]_Q[T]^{-1}_Q = I$ Let $(A)_i$ and $(A)^j$ denote the $i$-th row and $j$-th column of $A$ respectively. Then by the definition of matrix multiplication and Hermititan transpose $[T]_Q{[T^*]_Q}_{ij} = \langle ([T]_Q)^i, {([T]^*_Q)}_j \rangle = \langle ([T]_Q)^i, {([T]_Q)}^j \rangle = \delta_{ij}$ So the columns of $[T]_Q$ are orthonormal.