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I am trying to prove the following proposition:

If $f$ is an endomorphism of an Euclidean space $(E,\langle \cdot, \cdot \rangle)$, then TFAE:

  • $f$ is an isometry of $E$.

  • The representative matrix of $f$ relative to any orthonormal basis of $E$ is orthogonal.

  • There exists an orthonormal basis of $E$ for which the representative matrix of $f$ relative to this basis is orthogonal.

I'm stuck on how to prove that the last implies the first.

Proof of the others:

Let $\{u_1, \ldots, u_n\}$ be an orthonormal basis of $E$. As $f$ is an isometry, then $\{f(u_i), i\}$ is an orthonormal basis as well.

Let $A = (a_{ij})$ be the representative matrix of $f$ relative to the $u_i$'s. Then, $a_{ij} = \langle f(u_j), u_i \rangle$.

Let $B = A A^t = b_{ij}$. Then:

$$b_{ij} = \sum_{t = 1}^n a_{it}a_{jt} = \sum_{t=1}^n \langle f(u_t), u_i \rangle \langle f(u_t), u_j \rangle = \langle u_i, u_j \rangle = \delta_{ij}$$

Therefore $B=I_n$ and this completes the proof.

The other one is obvious.

Thanks for your help.

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  • $\begingroup$ < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. $\endgroup$ – Zev Chonoles Jun 13 '15 at 1:07
  • $\begingroup$ @ZevChonoles thanks. This looks much better. $\endgroup$ – user230734 Jun 13 '15 at 1:11
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Observe that if $\{e_i\}$ is an orthonormal basis, the respective Gram matrix representing $\langle \cdot, \cdot \rangle$ is $(\langle e_i, e_j \rangle)$, which is the identity matrix.

So $\forall v\in E$, $\langle fv , fv\rangle = v^tf^tfv= v^tv = \langle v , v\rangle$.

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