# Orthonormal basis - Set of all isometries

#### mathmari

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Hey!!

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:
1. Let $b_i:=\phi (e_i)$ for all $1\leq i\leq n$. Then $(b_1, \ldots , b_n)$ is an orthonormal basis.
2. For $v\in V$ it holds that $\displaystyle{\phi(v)=\sum_{i=1}^n(v\cdot e_i)b_i}$
3. For $v\in V$ it holds that $\phi (v)=(b_1\mid \ldots \mid b_n)v$.
4. There is a matrix $b\in O_n$ such that $\phi=\phi_b$.
5. 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?

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 $\displaystyle{\phi \left (\sum_{i=1}^n\left (v\cdot e_i\right )e_i\right )=\sum_{i=1}^n\left (v\cdot e_i\right )\phi (e_i )}$ ?

For 3:
Do we maybe use the result of the previous question? Is the following correct? $$\phi (v)=\sum_{i=1}^n\left (v\cdot e_i\right )b_i=\sum_{i=1}^nv \left (e_i\cdot b_i\right )$$

Could you give me a hint for 4. and 5. ?

#### Klaas van Aarsen

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

That would be a consequence of the fact that $\phi$ is a linear transformation, wouldn't it?

For 3:
Do we maybe use the result of the previous question?
Yes.
Consider what $(b_1\mid \ldots \mid b_n)v$ means.
I think it is the sum of the products of each $b_i$ with each component of $v$ with respect to the standard basis, isn't it?
Can we write it as an expression with a sum operator?

Is the following correct? $$\phi (v)=\sum_{i=1}^n\left (v\cdot e_i\right )b_i=\sum_{i=1}^nv \left (e_i\cdot b_i\right )$$
That doesn't look correct.
Generally matrix operations are associative, but those are not proper matrix operations are they?
The dimensions do not match do they?

Could you give me a hint for 4. and 5. ?
For 4, suppose we pick the matrix with entries $(b_i\cdot e_j)$.
Does it match the desired $\phi$?

For 5, can we write each isometry that preserves the origin as a matrix with orthonormal columns?
Does each matrix with orthonormal columns have an associated transformation that is an isometry that preserves the origin?

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#### mathmari

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Consider what $(b_1\mid \ldots \mid b_n)v$ means.
I think it is the sum of each the products of each $b_i$ with each component of $v$ with respect to the standard basis, isn't it?
Can we write it as an expression with a sum operator?
Ahh I understood it now! We have that $v\cdot e_i$ is the $i$-th component of the vector $v$.
Then $$(b_1\mid \ldots \mid b_n)v=\sum_{i=1}^nb_i(v\cdot e_i)=\sum_{i=1}^n(v\cdot e_i)b_i$$

So to justify that it is like that is it complete and formallly to say that when we multiply a matrix and a vector then the result is a vector which is the sum of each column of the matrix multiplied by the respective component of the vector?

#### Klaas van Aarsen

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So to justify that it is like that is it complete and formallly to say that when we multiply a matrix and a vector then the result is a vector which is the sum of each column of the matrix multiplied by the respective component of the vector?
I believe so yes.

#### mathmari

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For 4, suppose we pick the matrix with entries $(b_i\cdot e_j)$.
Does it match the desired $\phi$?
If we have this matrix $b$ then we get that $\phi_b(v)=bv$. Then we use question 2 or 3 to get that this is equal to $\phi(v)$, right?
Are the entries that you mentioned the say if we say that the matrix is $b=(b_1\mid \ldots \mid b_n)$ and the result follows from the previous question? Or is this wrong?

#### Klaas van Aarsen

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If we have this matrix $b$ then we get that $\phi_b(v)=bv$. Then we use question 2 or 3 to get that this is equal to $\phi(v)$, right?
Are the entries that you mentioned the say if we say that the matrix is $b=(b_1\mid \ldots \mid b_n)$ and the result follows from the previous question? Or is this wrong?
Effectively yes.
As I understand it, the given $b=(b_1\mid \ldots \mid b_n)$ is a matrix that consists of 1 row where each element is an abstract vector.
The matrix given by $(b_i\cdot e_j)$ is an $n\times n$-matrix of real numbers that has each $b_i$ as a column vector with respect to the standard basis.
I think we're supposed to show that this matrix is an orthogonal matrix, which is a matrix with orthonormal columns. That's what $O_n$ represents, isn't it?

#### mathmari

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As I understand it, the given $b=(b_1\mid \ldots \mid b_n)$ is a matrix that consists of 1 row where each element is an abstract vector.
But $(b_1, \ldots , b_n)$ is an orthonormal basis, according to question 1, and so these are not abstract vectors, or do you mean something else?

#### Klaas van Aarsen

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But $(b_1, \ldots , b_n)$ is an orthonormal basis, according to question 1, and so these are not abstract vectors, or do you mean something else?
Isn't a vector an abstract object that has multiple representations?
The representation depends on the basis we choose, doesn't it?
As long as we don't select a basis, isn't a vector just an abstract object?

#### mathmari

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Isn't a vector an abstract object that has multiple representations?
The representation depends on the basis we choose, doesn't it?
As long as we don't select a basis, isn't a vector just an abstract object?
I got stuck right now. Isn't the vector of question 1 a basis? Or am I thinking wrong?

#### Klaas van Aarsen

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I got stuck right now. Isn't the vector of question 1 a basis? Or am I thinking wrong?
Yes... the vectors $b_i$ form a basis... so?

What do you think we should do for the question?

#### mathmari

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Let $v\in V$ then let $b$ a matrix then we have that $\phi_b(v)=bv$ and we have from question 3 using the vectors $b_i=\phi(e_i)$ (which form an orthonormal basis) that $\phi (v)=(b_1\mid \ldots \mid b_n)v$. Therefore we could use the matrix $b=(b_1\mid \ldots \mid b_n)=(\phi(e_1)\mid \ldots \mid \phi(e_n))$ to get $\phi_b(v)=\phi (v)$.
The matrix $b$ is then orthogonal since the columns form an orthonormal basis.

#### Klaas van Aarsen

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Let $v\in V$ then let $b$ a matrix then we have that $\phi_b(v)=bv$ and we have from question 3 using the vectors $b_i=\phi(e_i)$ (which form an orthonormal basis) that $\phi (v)=(b_1\mid \ldots \mid b_n)v$. Therefore we could use the matrix $b=(b_1\mid \ldots \mid b_n)=(\phi(e_1)\mid \ldots \mid \phi(e_n))$ to get $\phi_b(v)=\phi (v)$.
The matrix $b$ is then orthogonal since the columns form an orthonormal basis.
Looks good to me.

#### mathmari

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Great!!

Could you explain to me further your hint for question 5?

#### Klaas van Aarsen

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Could you explain to me further your hint for question 5?
We have to show that each element in the left hand set is also an element in the right hand set, don't we?
And vice versa as well.

On the left we have an isometry that preserves the origin.
And on the right we have a map given by an orthogonal matrix don't we?

#### mathmari

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We consider that $T(v)=Av$ is an isometry. Do we have then the following?
\begin{align*}\|x-y\|=\|Ax-Ay\|&\iff \langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle \\ & \iff \langle x, y\rangle=\langle Ax, Ay\rangle \\ & \iff \langle x, y\rangle=\langle x, A^TAy\rangle\\ & \iff \langle x, A^TAy-y\rangle =0 \\ & \iff \langle x, (A^TA-I)y\rangle =0 \\ & \iff (A^TA-I)y =0 \\ & \iff A^TA-I =0 \\ & \iff A^TA=I \\ & \iff A\in O_n\end{align*} Is it correct to use everywhere $\iff$ ? Do we have then shown noth directions?

#### Klaas van Aarsen

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We consider that $T(v)=Av$ is an isometry. Do we have then the following?
\begin{align*}\|x-y\|=\|Ax-Ay\|&\iff \langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle \end{align*}
Interesting that you bring this up now.
Formally an isometry is a map $\phi$ such that for any x and y we have $d(\phi(x),\phi(y))=d(x,y)$.
It's from Greek where isometry means 'equal distance', isn't it?
With the default distance in a normed space that becomes $\|\phi(x)-\phi(y)\|=\|x-y\|$.
And with the default norm in an inner product space it becomes $\langle \phi(x)-\phi(y),\phi(x)-\phi(y)\rangle = \langle x-y,x-y\rangle$.

But you used $\phi(x)\cdot \phi(y)=x\cdot y$ in question 1.
I let it slide since that is correct for an isometry that preserves the origin, although that should probably be mentioned explicitly.

Anyway, an isometry cannot generally be written as $T(v)=Av$, so there is a hidden assumption in there.
Then the reasoning cannot be correct can it?

\begin{align*}\langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle \iff \langle x, y\rangle=\langle Ax, Ay\rangle\end{align*}
Is it correct to use everywhere $\iff$ ? Do we have then shown noth directions?
How did you get this?

#### mathmari

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Now I thought about that again. Can we not use here the previous question? We have that $\phi=\phi_b$ where $\phi\in O(V)$ and $b\in O_n$.

#### Klaas van Aarsen

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Now I thought about that again. Can we not use here the previous question? We have that $\phi=\phi_b$ where $\phi\in O(V)$ and $b\in O_n$.
Yep. That works.

That leaves us to show the converse doesn't it?

#### mathmari

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With this we show that $O(V) \subseteq \{\phi_a\mid a\in O_n\}$ or not? What do we have to show for the other direction?

#### Klaas van Aarsen

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With this we show that $O(V) \subseteq \{\phi_a\mid a\in O_n\}$ or not? What do we have to show for the other direction?
Yes.
For the other direction, suppose we have an $a\in O_n$, is $\phi_a$ then an isometry that preserves the origin?

#### mathmari

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For the other direction, suppose we have an $a\in O_n$, is $\phi_a$ then an isometry that preserves the origin?
We have that $\phi_a(v)=av$. Then we get \begin{align*}\| \phi_a(x)-\phi_a(y)\|&=\langle \phi_a(x)-\phi_a(y),\phi_a(x)-\phi_a(y)\rangle \\ & =\langle ax-ay,ax-ay\rangle \\ & =(ax-ay)^T(ax-ay)\\ & =(x^Ta^T-y^Ta^T)(ax-ay)\\ & =(x^T-y^T)a^T(ax-ay) \\ & =(x^T-y^T)(a^Tax-a^Tay)\\ & \ \overset{a\in O_n}{=}(x^T-y^T)(x-y)\\ & =(x-y)^T(x-y) \\ & = \langle x-y,x-y\rangle \\ & =\|x-y\|\end{align*} Which means that $\phi_a$ is an isometry and since $\phi_a(0)=a\cdot 0=0$ it is an isometry that preserves the origin.

Is everything correct?

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#### Klaas van Aarsen

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We have that $\phi_a(v)=av$. Then we get \begin{align*}\| \phi_a(x)-\phi_a(y)\|&=\langle \phi_a(x)-\phi_a(y),\phi_a(x)-\phi_a(y)\rangle \\ & =\langle ax-ay,ax-ay\rangle \\ & =(ax-ay)^T(ax-ay)\\ & =(x^Ta^T-y^Ta^T)(ax-ay)\\ & =(x^T-y^T)a^T(ax-ay) \\ & =(x^T-y^T)(a^Tax-a^Tay)\\ & \ \overset{a\in O_n}{=}(x^T-y^T)(x-y)\\ & =(x-y)^T(x-y) \\ & = \langle x-y,x-y\rangle\end{align*} Which means that $\phi_a$ is an isometry and since $\phi_a(0)=a\cdot 0=0$ it is an isometry that preserves the origin.

Is everything correct?
Yep.
Btw, you are using the property that a matrix $a$ with orthonormal columns has $a^Ta=I$ don't you?
And strictly speaking we should end with $\|x-y\|$ shouldn't we?

#### mathmari

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Btw, you are using the property that a matrix $a$ with orthonormal columns has $a^Ta=I$ don't you?
Yes, this correct, isn't it?

And strictly speaking we should end with $\|x-y\|$ shouldn't we?
As it is, you specified $a\in O_n$ as a rationale to use that $a^Ta=I$, but the latter isn't immediately evident from the fact that $a$ has orthonormal columns is it?