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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? –  user38268 Oct 29 '11 at 0:44

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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I assumed real base field, but the complex case shouldn't be much more difficult. –  Srivatsan Oct 29 '11 at 0:47
    
I am making no assumptions about the base field. It could be real, complex, finite, etc. I have shown now that the columns are orthogonal, but how can I show they are still length 1? I only used your first step to show orthogonality, and I don't really follow the other steps. –  nullUser Oct 29 '11 at 0:56
    
Did you understand what $c_i$ means? Where exactly are you stuck in (2.)? –  Srivatsan Oct 29 '11 at 0:59
    
I have shown that $[\ell(\vec{q_i})]_Q = \sum_j \langle \ell(\vec{q_i}),\vec{q_j} \rangle \vec{q_j}$ if that helps. –  nullUser Oct 29 '11 at 1:05
    
@Kb100 the base field must be either real or complex. Otherwise the inner product doesn't make sense. –  user12014 Oct 29 '11 at 1:25

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.

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