# [SOLVED]Orthogonal Transformation in Euclidean Space

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

Here's one of the questions that I encountered recently along with my answer. Let me know if you see any mistakes. I would really appreciate any comments, shorter methods etc.

Problem:

Let $$u,\,v$$ be two vectors in a Euclidean space $$V$$ such that $$|u|=|v|$$. Prove that there is an orthogonal transformation $$f:\, V\rightarrow V$$ such that $$v=f(u)$$.

Solution:

We assume that $$u$$ and $$v$$ are non zero. Otherwise the result holds trivially.

Let $$B$$ denote the associated symmetric bilinear function of the Euclidean space. Let us define the linear transformation $$f$$ as,

$f(x)=\begin{cases}x&\mbox{if}&x\neq u\\v&\mbox{if}&x=u\end{cases}$

It's clear that, $$B(f(x),\,f(y))=B(x,\,y)$$ whenever $$x,\,y\neq u$$. Also $$B(f(u),\,f(u))=B(v,\,v)$$ and since $$|v|=|u|\Rightarrow B(v,\,v)=B(u,\,u)$$ we have $$B(f(u),\,f(u))=B(u,\,u)$$.

It remains to show that, $$B(f(x),\,f(u))=B(x,\,u)$$ for $$x\neq u$$.

$B(f(v+u),\,f(v+u))=B(f(v),\,f(v))+2B(f(v),\,f(u))+B(f(u),\,f(u))$

Also since $$v+u\neq u$$,

$B(f(v+u),\,f(v+u))=B(v+u,\,v+u)=B(v,\,v)+2B(v,\,u)+B(u,\,u)$

Using the above two results and the fact that $$B(u,\,u)=B(v,\,v)$$ we get,

$B(f(v),\,f(u))=B(v,\,u)$

Now consider $$B(f(x+v),\,f(x+u))$$.

Case I: $$x+v\neq u$$

$B(f(x+v),\,f(x+u))=B(f(x),\,f(x))+B(f(x),\,f(u))+B(f(v),\,f(x))+B(f(v),\,f(u))$

Also,

$B(f(x+v),\,f(x+u))=B(x+v,\,x+u)=B(x,\,x)+B(x,\,u)+B(x,\,v)+B(v,\,u)$

Using the above two results and the fact that $$B(f(v),\,f(u))=B(v,\,u)$$ we get,

$B(f(x),\,f(u))=B(x,\,u)$

Case II: $$x+v=u$$

$B(x,\,v)=B(u-v,\,v)=B(u,\,v)-B(v,\,v)$
$B(x,\,u)=B(u-v,\,u)=B(u,\,u)-B(v,\,u)$
Therefore, $B(x,\,u)=-B(x,\,v)~~~~~~(1)$

$B(f(x),\,f(u))=B(f(u-v),\,f(u))=B(f(u),\,f(u))-B(f(v),\,f(u))=B(v,\,v)-B(v,\,v)=0$

Then since $$B(f(x),\,f(u))=B(x,\,v)=0$$ by (1) we get $$B(x,\,u)=0$$

$\therefore B(f(x),\,f(u))=B(x,\,u)$

#### Opalg

##### MHB Oldtimer
Staff member
Problem:

Let $$u,\,v$$ be two vectors in a Euclidean space $$V$$ such that $$|u|=|v|$$. Prove that there is an orthogonal transformation $$f:\, V\rightarrow V$$ such that $$v=f(u)$$.

Solution:

We assume that $$u$$ and $$v$$ are non zero. Otherwise the result holds trivially.

Let $$B$$ denote the associated symmetric bilinear function of the Euclidean space. Let us define the linear transformation $$f$$ as,

$f(x)=\begin{cases}x&\mbox{if}&x\neq u\\v&\mbox{if}&x=u\end{cases}$
The problem with this is that the map $f$ can never be linear (unless $u=v$).

It may help to think in terms of a simple example. In the space $V=\mathbb{R}^2$, let $u=(1,0)$ and $v=(0,1)$. The only orthogonal transformations taking $u$ to $v$ are a rotation of the whole space through a right angle, or a reflection of the whole space in the line $y=x$. Either way, the transformation has to shift just about every vector in the space. The map that just takes $u$ to $v$ and leaves everything else fixed is not linear, and certainly not orthogonal.

To prove this result, you need to construct a linear map $f$ taking $u$ to $v$. The way to do that is to define $f$ on an orthonormal basis for $V$ and then extend it by linearity to a map on the whole of $V$. Start by constructing an orthonormal basis $\{e_1,e_2,\ldots,e_n\}$ such that $e_1$ is a multiple of $u$. Then do the same for $v$, showing that there is an orthonormal basis $\{g_1,g_2,\ldots,g_n\}$ such that $g_1$ is a multiple of $v$. You can then define $f$ by $f(e_k) = g_k$ for $1\leqslant k\leqslant n$.

It should then be straightforward to check that the map $f$ is orthogonal.

#### Sudharaka

##### Well-known member
MHB Math Helper
The problem with this is that the map $f$ can never be linear (unless $u=v$).

It may help to think in terms of a simple example. In the space $V=\mathbb{R}^2$, let $u=(1,0)$ and $v=(0,1)$. The only orthogonal transformations taking $u$ to $v$ are a rotation of the whole space through a right angle, or a reflection of the whole space in the line $y=x$. Either way, the transformation has to shift just about every vector in the space. The map that just takes $u$ to $v$ and leaves everything else fixed is not linear, and certainly not orthogonal.

To prove this result, you need to construct a linear map $f$ taking $u$ to $v$. The way to do that is to define $f$ on an orthonormal basis for $V$ and then extend it by linearity to a map on the whole of $V$. Start by constructing an orthonormal basis $\{e_1,e_2,\ldots,e_n\}$ such that $e_1$ is a multiple of $u$. Then do the same for $v$, showing that there is an orthonormal basis $\{g_1,g_2,\ldots,g_n\}$ such that $g_1$ is a multiple of $v$. You can then define $f$ by $f(e_k) = g_k$ for $1\leqslant k\leqslant n$.

It should then be straightforward to check that the map $f$ is orthogonal.
Thanks so much for the informative reply. I think I am getting the idea. First we can choose an orthonormal basis $$\{e_1,e_2,\ldots,e_n\}$$ such that $$e_1$$ is a multiple of $$u$$. Then if we rotate this basis by a certain angle so as to align $$e_1$$ with $$v$$ we could get the basis $$\{g_1,g_2,\ldots,g_n\}$$. Since $$|v|=|u|$$ our new basis would have $$g_1$$ a multiple of $$v$$. Am I correct? Or is there a more formal way of doing this?

#### Deveno

##### Well-known member
MHB Math Scholar
I think what Opalg is getting at is this:

If you have a linear map that takes a basis to a basis, it is certainly invertible.

For $e_1$, we can always choose $e_1 = u/|u|$, and use something like Gram-Schmidt to turn any basis extension we create into an orthogonal basis:

$\{e_1,\dots,e_n\}$.

The same process is then used to create the 2nd basis:

$\{g_1,\dots,g_n\}$, where $g_1 = v/|v|$.

We then DEFINE, for any $x \in V$:

$T(x) = T(c_1e_1 + \cdots + c_ne_n) = c_1g_1 + \cdots + c_ng_n$.

Note that $T(u) = T(|u|e_1) = |u|T(e_1) = |v|T(e_1)$ (since $|u| = |v|$)

$= |v|g_1 = |v|(v/|v|) = v$.

Now proving orthogonality is a bit of a mess to write explicitly, but the idea is this:

Since both bases are ORTHOGONAL (we can actually insist on orthonormal by scaling the basis vectors to unit vectors), we have:

$B(e_i,e_i) = B(g_i,g_i) = 1$
$B(e_i,e_j) = B(g_i,g_j) = 0,\ i \neq j$.

So if:

$x = c_1e1 + \cdots + c_ne_n$
$y = d_1e_1 + \cdots + d_ne_n$, then:

$B(x,y) = B(c_1e1 + \cdots + c_ne_n,d_1e_1 + \cdots + d_ne_n)$

$\displaystyle = \sum_{i,j} c_id_jB(e_i,e_j) = \sum_i c_id_i$

by the bilinearity of $B$ and the orthogonality of our basis.

Similarly, evaluating $B(T(x),T(y))$ gives the same answer, and there ya go.

#### Deveno

##### Well-known member
MHB Math Scholar
Forgive the double-post, but I thought I would give an explicit example for $\Bbb R^3$.

For our symmetric Euclidean bilinear form, I will use the standard dot-product. It is possible to use so-called "weighted" inner products, but they just add needlessly complicated calculation to the scenario.

For our first vector, we will take $u = (2,0,0)$. For the second, we will take $v = (1,\sqrt{2},1)$, which I think are perfectly reasonable choices.

For our first basis, the usual $\{(1,0,0),(0,1,0),(0,0,1)\}$ will do quite nicely. The second basis is a bit of a pain to come up with, we start with the unit vector:

$g_1 = (\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})$.

To get a basis, we'll just add in (0,1,0) and (0,0,1) and apply Gram-Schmidt:

First, we calculate:

$(0,1,0) - \frac{(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})\cdot(0,1,0)}{(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})\cdot(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})}(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})$

$= (\frac{-\sqrt{2}}{4},\frac{1}{2},\frac{-\sqrt{2}}{4})$

and normalizing this gives us:

$g_2 = (\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})$

Finally, we calculate:

$(0,0,1) - \frac{(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})\cdot(0,0,1)}{(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})\cdot(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})}(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2}) - \frac{(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})\cdot(0,0,1)}{(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})\cdot(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})}(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})$

$= (\frac{-1}{2},0,\frac{1}{2})$ which upon normalization gives us:

$g_3 = (\frac{-\sqrt{2}}{2},0,\frac{\sqrt{2}}{2})$.

It is clear, then, then that the orthogonal linear mapping we are looking for is given by the matrix (relative to the standard basis for $\Bbb R^3$):

$[T] = \begin{bmatrix}\frac{1}{2}&\frac{-1}{2}&\frac{-\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}&0\\ \frac{1}{2}&\frac{-1}{2}&\frac{\sqrt{2}}{2} \end{bmatrix}$

which obviously (heh!) has determinant 1, and is orthogonal, and moreover:

$T(u) = T(2,0,0) = (1,\sqrt{2},1) = v$.

#### Sudharaka

##### Well-known member
MHB Math Helper
I think what Opalg is getting at is this:

If you have a linear map that takes a basis to a basis, it is certainly invertible.

For $e_1$, we can always choose $e_1 = u/|u|$, and use something like Gram-Schmidt to turn any basis extension we create into an orthogonal basis:

$\{e_1,\dots,e_n\}$.

The same process is then used to create the 2nd basis:

$\{g_1,\dots,g_n\}$, where $g_1 = v/|v|$.

We then DEFINE, for any $x \in V$:

$T(x) = T(c_1e_1 + \cdots + c_ne_n) = c_1g_1 + \cdots + c_ng_n$.

Note that $T(u) = T(|u|e_1) = |u|T(e_1) = |v|T(e_1)$ (since $|u| = |v|$)

$= |v|g_1 = |v|(v/|v|) = v$.

Now proving orthogonality is a bit of a mess to write explicitly, but the idea is this:

Since both bases are ORTHOGONAL (we can actually insist on orthonormal by scaling the basis vectors to unit vectors), we have:

$B(e_i,e_i) = B(g_i,g_i) = 1$
$B(e_i,e_j) = B(g_i,g_j) = 0,\ i \neq j$.

So if:

$x = c_1e1 + \cdots + c_ne_n$
$y = d_1e_1 + \cdots + d_ne_n$, then:

$B(x,y) = B(c_1e1 + \cdots + c_ne_n,d_1e_1 + \cdots + d_ne_n)$

$\displaystyle = \sum_{i,j} c_id_jB(e_i,e_j) = \sum_i c_id_i$

by the bilinearity of $B$ and the orthogonality of our basis.

Similarly, evaluating $B(T(x),T(y))$ gives the same answer, and there ya go.
Forgive the double-post, but I thought I would give an explicit example for $\Bbb R^3$.

For our symmetric Euclidean bilinear form, I will use the standard dot-product. It is possible to use so-called "weighted" inner products, but they just add needlessly complicated calculation to the scenario.

For our first vector, we will take $u = (2,0,0)$. For the second, we will take $v = (1,\sqrt{2},1)$, which I think are perfectly reasonable choices.

For our first basis, the usual $\{(1,0,0),(0,1,0),(0,0,1)\}$ will do quite nicely. The second basis is a bit of a pain to come up with, we start with the unit vector:

$g_1 = (\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})$.

To get a basis, we'll just add in (0,1,0) and (0,0,1) and apply Gram-Schmidt:

First, we calculate:

$(0,1,0) - \frac{(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})\cdot(0,1,0)}{(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})\cdot(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})}(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})$

$= (\frac{-\sqrt{2}}{4},\frac{1}{2},\frac{-\sqrt{2}}{4})$

and normalizing this gives us:

$g_2 = (\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})$

Finally, we calculate:

$(0,0,1) - \frac{(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2})\cdot(0,0,1)}{(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})\cdot(\frac{1}{2},\frac{\sqrt{2}}{2}, \frac{1}{2})}(\frac{1}{2},\frac{\sqrt{2}}{2},\frac{1}{2}) - \frac{(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})\cdot(0,0,1)}{(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})\cdot(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})}(\frac{-1}{2},\frac{\sqrt{2}}{2},\frac{-1}{2})$

$= (\frac{-1}{2},0,\frac{1}{2})$ which upon normalization gives us:

$g_3 = (\frac{-\sqrt{2}}{2},0,\frac{\sqrt{2}}{2})$.

It is clear, then, then that the orthogonal linear mapping we are looking for is given by the matrix (relative to the standard basis for $\Bbb R^3$):

$[T] = \begin{bmatrix}\frac{1}{2}&\frac{-1}{2}&\frac{-\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}&0\\ \frac{1}{2}&\frac{-1}{2}&\frac{\sqrt{2}}{2} \end{bmatrix}$

which obviously (heh!) has determinant 1, and is orthogonal, and moreover:

$T(u) = T(2,0,0) = (1,\sqrt{2},1) = v$.
Hi Denevo,

Thanks very much for both of your posts. After reading them I understood almost everything that is required to solve the problem. Now I think I should read more about the Gram-Schimdt process.