Proof about isometries: V=R^n furnished w/ standard inner product, standard basis S.

TheSky

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Jun 21, 2016
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Hello,

i'm trying to prove this statements, but i'm stuck.

Be \(\displaystyle V=R^n\) furnished with the standard inner product and the standard basis S.
And let W \(\displaystyle \subseteq\) V be a subspace of V and let \(\displaystyle W^\bot\) be the orthogonal complement.

a) Show that there is exactly one linear map \(\displaystyle \Phi:V \rightarrow V\) with \(\displaystyle \Phi|_w=id_w\) and with \(\displaystyle \Phi|_{w^\bot}=-id_{w^\bot}\)

b) Show that V have an orthonormal basis B consisting of the eigenvectors of \(\displaystyle \Phi\) and indicate \(\displaystyle D_{BB}(\Phi\)

c) Show that \(\displaystyle D_{BS}(id_v)\) and \(\displaystyle D_{SS}(\Phi)\) are orthogonal matrices.




For a) i have the following incomplete derivation:

Be \(\displaystyle a_1\)...\(\displaystyle a_n\) an orthonormal basis of W and be \(\displaystyle b_1\)...\(\displaystyle b_n\) an orthonormal basis of \(\displaystyle W^\bot\).
Therefore \(\displaystyle \Phi\) is defined as \(\displaystyle \Phi: a_i \mapsto a_i, b_i \mapsto -b_j\) with 1\(\displaystyle \le\)i\(\displaystyle \le\)n and 1\(\displaystyle \le\)j\(\displaystyle \le\)n. We can see that \(\displaystyle a_i\) and \(\displaystyle b_i\) are eigenvectors of \(\displaystyle \Phi\).

And now i'm stuck. I'm sure, i saw somewhere an prove with this derivation. But i dont remember. Is this even a good starting point or a dead end?
Well, i'm not very good at mathematical prooves.
But maybe someone can help me with the next step or someone have an other idea to proove this.
Thanks in advance.
 
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