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Let $L$ be a transformation on a finite-dimensional real vector space $V$ with an inner product. Also $L$ sends each orthonormal basis in $V$ to another orthonormal basis. Prove that $L$ is an orthogonal transformation.

I have trouble proving the first step which is trying to prove that this transformation is linear. How should I continue? Any hints?

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  • $\begingroup$ If $L$ is a linear transformation $V\to V$ which sends orthonormal bases to orthonormal bases, and $f:[0,\infty)\to[0,\infty)$ is any scalar function with $f(0)=0$ and $f(1)=1$, then $v\mapsto f(|v|)L(v)$ is also a function $V\to V$ which sends orthonormal bases to orthonormal bases but may not be a linear transformation. This suggests you should interpret "transformation" to automatically mean linear transformation. It would be an interesting question to prove it's linear assuming it satisfied the condition $L(rv)=rL(v)$ though (either for all $r$ or for $r\ge0$). $\endgroup$
    – anon
    Dec 3, 2017 at 17:15
  • $\begingroup$ The case when $L$ is a linear transformation is as you said very easy to prove,but I do not think I can assume that. $\endgroup$ Dec 4, 2017 at 3:25
  • $\begingroup$ I did not say it was "very easy to prove" when $L$ is assumed linear. I was showing how the conclusion is false if we don'e assume something about $L$ - namely, that it commutes with scaling lengths of vectors. $\endgroup$
    – anon
    Dec 4, 2017 at 5:38

2 Answers 2

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Let $v=\sum v_i e_i$ (basically a basis expansion of v). Then: $Lv = L(\sum v_i e_i)$. Use the IP against some $e_j$: $$\left\langle L(\sum v_i e_i), e_j\right\rangle \\ = \left\langle \sum v_i e_i, L^*(e_j)\right\rangle \text{adjoint always exists in finite dims} \\ = \sum v_i \left\langle e_i, L^*(e_j)\right\rangle \text{linearity of IP} \\ = \sum v_i \left\langle L(e_i), e_j\right\rangle \\ = \sum \left\langle v_i L( e_i), e_j\right\rangle \\ = \left\langle \sum v_i L( e_i), e_j\right\rangle $$

This is true for all $e_j$ so $L\left(\sum v_i e_i\right) = \sum v_i L( e_i)$

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  • $\begingroup$ Adjoint always exist in finite dimension? Isn't it true after assuming $L$ to be linear. $\endgroup$
    – Mr. X
    Dec 3, 2017 at 17:00
  • $\begingroup$ Doesn't just boundedness suffice? I'm hunting around now $\endgroup$
    – yoshi
    Dec 3, 2017 at 17:08
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    $\begingroup$ I had never heard of adjoint of any non linear map. $\endgroup$
    – Mr. X
    Dec 3, 2017 at 17:12
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Let be $A = \{e_1,...,e_n\}$ ortonormal base of $V$. By hipotesis $B =\{Le_1,...Le_n\}$ it's an ortonormal base of $V$, therefore, if $a_{ij} \in R$ are such that

$$Te_j = \sum_{i=1}^{n}a_{ij}e_i$$

we have from ortogonalyties of $A$ and $B$ that

$$\delta_{jk} = \langle Te_j,Te_k \rangle = \sum_{i = 1}^n a_{ij}a_{ik} = \sum_{i = 1}^n a_{ki}^Ta_{ij} $$

and therefore, $(a_{ki}^T)\times (a_{jl}) = I$, where $(a_{ki}^T)$ is transpost matrix of $(a_{jl})$, $I$ identity matriz of $n\times n$ ordem. Because $(a_{jl})$ is matrix of $L$ on terms of $A$, we have $L$ is ortogonal.

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  • $\begingroup$ OP is asking how to prove $L$ is linear. $\endgroup$
    – anon
    Dec 3, 2017 at 17:20

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