How should I start of? I know if $T$ is an isometry then $T=A \vec x+\vec b$ where A is orthogonal matrix and $\vec b$ is a column vector. If I can prove that there is no orthogonal matrix from $\mathbb{R^3}$ to $\mathbb{R^2}$ then will it suffice?(which I don't think is true?)
(Also my understanding of isometries is quite basic, I know isometries preserve distance and dot product)
Prove that there is no isometry from $\mathbb R^3$ to $\mathbb R^2$
geometry
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Ok, here is my idea, I hope this is what you had in mind:
We use one reflection to get an isometry with $f(0)=0$. I will now assume that $f$ has this property and will show that $n$ reflections suffice. edit: I should probably say at this point that the reason for step 1 is to use that $f$ is now norm-preserving (and thereby distance preserving) which simplifies the notation.
I will inductively construct reflections $r_i$ such that their composition $r_1\circ\cdots\circ r_n$ is the inverse of $f$. Since reflections are their own inverse this proves the statement.
Assume that $f$ fixes a subspace $N$ of dimension $d$ (and this subspace is maximal in the sense that there exists no higher dimensional subspace which is fixed). We may then write $\mathbb R^n=N\bot N^\bot$. I will show that we can find a reflection $r$, such that $r\circ f$ fixes a subspace of one dimension higher.
Take $0\neq x\in N^\bot$. Then $x\neq f(x)\in N^\bot$ since $0=\left< x,n\right>=\left< f(x),f(n)\right>=\left< f(x),n\right>$ for all $n\in N$. Moreover $f(x)$ is either linearly independent to $x$ or just $-x$. Indeed assume $\lambda f(x)=x$, then $\left< x,x\right>=\left< f(x),f(x)\right>=\lambda^2\left< x,x\right>$ and $\lambda^2=1$ with $\lambda\neq 1$ by assumption. If $f(x)=-x$, we just reflect $x$ and keep $N$ fixed at the same time and are done. Hence assume $f(x)$ and $x$ are linear independent.
Find a basis of $N^\bot$ containing $x$ and $f(x)$ namely $\{x,f(x),b_1,...,b_k\}$. Choose $u=x+f(x)$ and $v=x-f(x)$. Then $\{u,v,b_1,...,b_k\}$ is also a basis of $N^\bot$, since $x=1/2(u+v)$ and $f(x)=1/2(u-v)$.
Take $r$ te be the reflection which takes $v$ to $-v$ and fixes everthing else. Then $r\circ f$ fixes the $d+1$-dimensional space $N\bot \mathbb Rx$.
After $n$ steps we have (at most) $n$ reflections such that $r_1\circ...\circ r_n\circ f$ is the identity. We are done.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$When you speak without qualification of an isometry with respect to the standard Hermitian norm, you're speaking of an isometry of Euclidean $4$-space. The story of Euclidean isometries is more-or-less uniform independently of dimension, so let's speak of Euclidean $n$-space with $n$ an arbitrary positive integer. Here are some useful factlets:
If an isometry $F$ fixes the origin and the standard basis vectors, then $F$ is the identity.
Sketch of proof: This is clear if $n = 1$. To argue inductively on the dimension, decompose a point of $\Reals^{m+1}$ as $(x, x_{m}) = ((x_{1}, \dots, x_{m}), x_{m+1})$, and show $F$ preserves both $\|x\|$ (so $F(x) = x$ by the inductive hypothesis) and the $(m + 1)$th coordinate $x_{m+1}$.
If an isometry $F$ fixes the origin, there exists a unique $n \times n$ (real) orthogonal matrix $A$ such that $F(x) = Ax$ for all $x$ in $\Reals^{n}$.
Sketch: By the isometry condition and the polarization identity $u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})$, the points $F(e_{i})$ constitute an orthonormal set in $\Reals^{n}$. Let $A$ be the unique orthogonal matrix satisfying $F(e_{i}) = Ae_{i}$ for each $i$, and apply the first item to $A^{-1} \circ F$. (This answers your first question affirmatively.)
If $F$ is an isometry, there exists a unique $n \times n$ orthogonal matrix and a unique $b$ in $\Reals^{n}$ such that $$ F(x) = Ax + b\quad\text{for all $x$ in $\Reals^{n}$.} $$ (You already have this proof. :)
If $p$ is a point of $\Reals^{n}$ and $u$ is a unit vector, then reflection in the hyperplane through $p$ orthogonal to $u$ is the mapping $$ R(x) = x - 2[(x - p) \cdot u] u, $$ namely, $x$ minus "twice the $u$-component of $x - p$". For example, if $u = e_{n}$ and $p$ is the origin, then $$ R(x_{1}, \dots, x_{n-1}, x_{n}) = (x_{1}, \dots, x_{n-1}, x_{n}) - 2x_{n} e_{n} = (x_{1}, \dots, x_{n-1}, -x_{n}). $$ The Euclidean group is generated by the set of reflections. In fact, if $F$ is a Euclidean isometry, there exist at most $(n + 1)$ reflections (not uniquely-defined!) whose composition is $F$.
A non-trivial translation is a composition of two reflections. (Why?)
A rotation is the composition of at most $n$ reflections, and is always the composition of an even number of reflections (because a reflection reverses orientation while a rotation preserves orientation). Thus, for example, every rotation of Euclidean $3$-space is a composition of at most two reflections, while a rotation of Euclidean $4$- or $5$-space is a composition of at most four reflections. (To prove this inductively, note that if $u$ and $v$ are arbitrary unit vectors, there exists a reflection exchanging $u$ and $v$.)
If, on the other hand, you want to speak only of Euclidean isometries of $\Cpx^{m} \simeq \Reals^{2m}$ for which the "rotation part" is complex-linear, you have the much smaller "holomorphic isometry group" corresponding to the unitary group $U(m) \subset SO(2m)$.
Thinking of $\Cpx^{2}$, the set of complex lines (a special class of real $2$-plane) is the well-known complex projective line, topologically a $2$-sphere, while the set of (oriented) real $2$-planes is topologically a product $S^{2} \times S^{2}$. This shows geometrically how small $U(2)$ is compared to $SO(4)$,
Best Answer
Consider a tetrahedron in $\mathbb R^3$. Its vertices are four points equidistant from each other.
An isometry with $\mathbb R^2$ would create such a set in $\mathbb R^2$, but this is clearly impossible.
One way to see is that since distinct circles in $\mathbb R^2$ only intersect in at most two locations. If you pick two points $A,B$ distance $d$ apart, and you want to find two more points distance $d$ away from $A,B$, then you must select the intersections $C,D$ of circles radius $d$ around $A,B$. But unfortunately, $C$ and $D$ are necessarily $d\sqrt{3}$ apart, not $d$!