Proving Isometries: A Step-By-Step Guide

[Doradus]

Hello,

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

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

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

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

c) Show that $D_{BS}(id_v)$ and $D_{SS}(\Phi)$ are orthogonal matrices.
For a) i have the following incomplete derivation:

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

And now I'm stuck. I'm sure, i saw somewhere an prove with this derivation. But i don't 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.

 

[RUber]

I am somewhat unfamiliar with your notation.
Could you please provide a bit more detail about what is meant by:
$\Phi |_\omega, d_\omega, D_{BB}, D_{BS}, \text{ and } D_{SS}$

One thing I notice right off is you define two n-dimensional basis sets -- one spanning W and the other spanning Wperp. With n vectors, you should span both spaces, or all of V.
Let $\{ a_i\}_{i=1}^n$ be a basis set for V, ordered in such a way that $\exists k,$ such that $\{ a_i\}_{i=1}^k$ is a basis set for $W$ and $\{ a_i\}_{i=k+1}^n$ is a basis set for $W^\perp$.

 

[Doradus]

$\Phi |_W$ is the same as $\Phi(W)$
$id_W$ is the identity funktion $\Phi(w)=w$
$D_{BB}$ Matrix with basis B
$D_{SS}$ Matrix with basis S
$D_{BS}$ I am not sure. :-)

Well, because I'm not sure, what $D_{BS}$ means, i think c) is not that important. I'm more interested in a) and b).

 

[RUber]

I see. thanks for the explanation.
For the first one, assume there are two linear maps then show that they must be equal. Because a linear map can be uniquely defined by its matrix representation, showing that the matrix representation must be the same should work.
$D_{BB}$ is a matrix that takes an input from basis B and gives an output in basis B.
$D_{SS}$ is a matrix that takes an input from basis S and gives an output in basis S.
Then, $D_{BS}$ should be a matrix that takes an input from basis B and gives an output in basis S.

Look at a simple example, Let $V = \mathbb{R}^3$, then $S = \{ \hat x, \hat y, \hat z\}$, $W$ is the xy-plane. $W^\perp$ is span of $\hat z$.
The matrix representation of $[\Phi ]_{SS} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
In your question 2, you are asked to give the matrix representation in the eigenbasis B...which should be pretty similar.