- #1

mathmari

Gold Member

MHB

- 5,049

- 7

Let $1\leq n\in \mathbb{N}$ and $\mathbb{R}^n$. A basis $B=(b_1, \ldots, b_n)$ of $V$ is an orthonormal basis, if $b_i\cdot b_j=\delta_{ij}$ for all $1\leq i,j,\leq n$.

Let $E=(e_1, \ldots,e_n)$ be the standard basis and let $\phi \in O(V)$. ($O(V)$ is the set of all isometries $\alpha$ such that $\alpha(0_V)=0_V$)

I want to show the following:

- Let $b_i:=\phi (e_i)$ for all $1\leq i\leq n$. Then $(b_1, \ldots , b_n)$ is an orthonormal basis.
- For $v\in V$ it holds that $\displaystyle{\phi(v)=\sum_{i=1}^n(v\cdot e_i)b_i}$
- For $v\in V$ it holds that $\phi (v)=(b_1\mid \ldots \mid b_n)v$.
- There is a matrix $b\in O_n$ such that $\phi=\phi_b$.
- It holds that $O(V)=\{\phi_a\mid a\in O_n\}$.

I have done the following:

**For 1:**

It holds that $\phi (x)\cdot \phi (y)=x\cdot y$ for all $x,y\in V$.

So we have that \begin{equation*}b_i\cdot b_j=\phi (e_i)\cdot \phi (e_j)=e_i\cdot e_j=\delta_{ij}=\begin{cases}1 , & i=j \\ 0 , & i\neq j\end{cases}\end{equation*}

We also have for all $1\leq j\leq n$ that \begin{equation*}\sum_{i=1}^n\lambda_ib_i=0 \Rightarrow

\left (\sum_{i=1}^n\lambda_ib_i\right )\cdot b_j=0\cdot b_j \Rightarrow

\sum_{i=1}^n\lambda_i\left (b_i\cdot b_j\right )=0\Rightarrow

\lambda_j=0 \end{equation*}

This means that $(b_1, \ldots , b_n)$ are linearly independent and this the number of these vectors is equal to the dimension of $V$, it follows that $(b_1, \ldots , b_n)$ is a basis of $V$.

For the basis $(b_1, \ldots , b_n)$ of $V$ it also holds that $b_i\cdot b_j=\delta_{ij}$ for all $1\leq i,j\leq n$, so it follows that $(b_1, \ldots , b_n)$ is an orthonormal basis.

Is this correct? :unsure:

**For 2:**

Since $(e_1, \ldots , e_n)$ is a absis of $V$, each element $v\in V$ can be written as:

\begin{equation*}v=\sum_{i=1}^n\lambda_ie_i \Rightarrow v\cdot e_j=\left (\sum_{i=1}^n\lambda_ie_i\right )\cdot e_j \Rightarrow v\cdot e_j=\sum_{i=1}^n\lambda_i\left (e_i\cdot e_j\right )\Rightarrow v\cdot e_j=\lambda_j, \ 1\leq j\leq n\end{equation*}

We get that \begin{equation*}v=\sum_{i=1}^n\left (v\cdot e_i\right )e_i \Rightarrow \phi (v)=\phi \left (\sum_{i=1}^n\left (v\cdot e_i\right )e_i\right ) \Rightarrow \phi (v)=\sum_{i=1}^n\left (v\cdot e_i\right )\phi (e_i ) \Rightarrow \phi (v)=\sum_{i=1}^n\left (v\cdot e_i\right )b_i\end{equation*}

Is it correct that $\displaystyle{\phi \left (\sum_{i=1}^n\left (v\cdot e_i\right )e_i\right )=\sum_{i=1}^n\left (v\cdot e_i\right )\phi (e_i )}$ ? :unsure:

**For 3:**

Do we maybe use the result of the previous question? Is the following correct? $$\phi (v)=\sum_{i=1}^n\left (v\cdot e_i\right )b_i=\sum_{i=1}^nv \left (e_i\cdot b_i\right )$$ :unsure:Could you give me a hint for 4. and 5. ? :unsure: