I am confused as to why an orthogonal transformation is necessarily invertible. I have defined an orthogonal transformation as a transformation $T:V\to V$ where $V$ is a vector space and $T$ has the property that $\langle T(x),T(y)\rangle=\langle x,y\rangle$ for all $x,y\in V$, and $\langle\cdot,\cdot\rangle$ is an arbitrary inner product. Any advice on how to prove this, or a general explanation would be much appreciated.
[Math] Why is an orthogonal transformation invertible
linear algebraorthogonality
Related Solutions
There are a couple ways to view a dot product as a linear map by changing your view slightly.
The map $\langle \cdot,\cdot \rangle : V\times V\to F$ is not linear, it is what we call bilinear, which means that it is linear in each variable. I.e. a map $B : V\times V' \to W$ is bilinear if for all fixed $v\in V$, and all fixed $v'\in V'$ the maps $u'\mapsto B(v, u')$ and $u\mapsto B(u, v')$ are linear. (Though these maps are equal in the case of a dot product, since it is symmetric, so you just need to check that one is linear. Symmetric meaning that $\langle v, w\rangle= \langle w, v\rangle$)
The other view is perhaps a little more faithful to the idea of viewing the dot product as a linear map, but essentially equivalent. Though perhaps a little more abstract (though such judgments are inherently subjective).
The idea is that we can take a bilinear map $B: V\times V' \to W$ and turn it into a linear map $\tilde{B}: V\to \newcommand\Hom{\operatorname{Hom}}\Hom_F(V',W)$. Where $\Hom_F(V',W)$ denotes the vector space of $F$-linear maps from $V'$ to $W$. We define $\tilde{B}(v) = v'\mapsto B(v,v')$. Then one can use the defining property of bilinear maps given above to show that $\tilde{B}$ is linear, and for any $v\in V$, $\tilde{B}(v)$ is a linear map from $V'$ to $W$. This process is called currying. Then $\tilde{B}$ is basically the same as $B$, since we can recover $B$ from $\tilde{B}$ from the fact that $B(v,v')=(\tilde{B}v)v'$ (sorry for changing notation to parenthesis-less function application, I just think it's much more readable here).
Thus one can curry the dot product to get a linear map, call it $D$ from $V$ to $\Hom_F(V,F)$. In general, $\Hom_F(V,F)$ is a vector space called $V$-dual, often written $V^*$, so we can say $D$ is a linear map from $V$ to $V^*$. I.e., we can view the dot product as being equivalent to a particular nice linear map from $V$ to $V^*$.
$a$ and $b$ are variable vectors and you have to find condition(s) on $T$ (not on these vectors) which make(s) $\langle a,b\rangle_1$ an inner product. Apply definition of inner product. $\langle a,b\rangle_1$ satisfies all properties of inner product for any $T$ except the condition $\langle a,a\rangle_1=0$ implies $a=0$. In other words what we need is $\|Tx\|=0$ implies $x=0$. This is true iff $T$ is injective.
Best Answer
For any vector $v\neq0$ the distance from o is positive. If An orthogonal transformation were singular then for a $v\neq0$ in it's kernel, $T(v)$ being zero would be at zero distance from the zero vector, contradicting the fact that it has to preserve the distance.