If $T:\mathbb{R^n} \to \mathbb{R^m}$ and $\langle Tx,Ty\rangle _m= \langle x,y\rangle _n$ then what can we say about $n$ and $m$? Let $\langle,\rangle$ denote the standard inner product on the vector space $\mathbb{R^n}$ such that $\langle x,y\rangle  = \sum_{i=1}^{n}x_iy_i$ for vectors $x,y \in R^n$. Let $T:\mathbb{R^n}\to \mathbb{R^m}$ be a linear map such that $\langle Tx,Ty\rangle _m= \langle x,y\rangle _n$. Then which of the following is necessarily true:$\\$
(a) $n \geq m$ (b) $n \leq m$ (c) $n=m$ (d) The map $T$ is onto.
Attempt: From the given condition we have $\langle Te_i, Te_j\rangle = \langle e_i,e_j\rangle = 1  $ If $i =j$ ,otherwise $0$. Let $ A$ be the matrix of $T$ with respect to the standard basis then $A^{T}A =I_n$ and $AA^{T} = I_m$ .So, $A$ is injective from the first equality and $A$ is surjective from the second inequality. Hence $\dim(R^n) = \dim (R^m) \implies n=m$
But the answer is $(b)$. So what is wrong with my approach?
 A: If you look at $T:\mathbb R\to\mathbb R^2, x\mapsto (x,0)$, you have the representation in the standard basis as
\begin{equation}
T(x)=\begin{pmatrix}1\\0\end{pmatrix} x
\end{equation}
and
\begin{equation}
\langle Tx, Ty\rangle = x y + 0 \cdot 0 = xy.
\end{equation}
Now,
\begin{equation}
\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=1=\mathrm{Id}_1,
\end{equation}
but
\begin{equation}
\begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}\neq\mathrm{Id}_2.
\end{equation}
In particular, this gives a counter-example to (c), (a) and (d).

(b) can be proved as follows: $T$ must be injective, because if $x\neq y$, then $$\lvert T(x-y)\rvert^2=\langle T(x-y),T(x-y)\rangle=\langle x-y, x-y\rangle\neq 0,$$ so that $Tx\neq Ty$.

The previous argument in your formulation: For any $x,y\in\mathbb R^n$, we have $$\langle x,y\rangle = \langle T x, T y\rangle = \langle A x, A y\rangle = \langle x, A^\top A y\rangle.$$ Inserting the basis vectors for $x$ and $y$ implies that the linear map $A^\top A$ must be the identity. There is no way to conclude that $AA^\top$ is the identity, though.
