Properties of length preserving linear transformations Let $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a linear transformation that preserves length. That is $||T(x)||=||x||$ for all $x\in\mathbb{R}^n$.Then: 


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*If $\langle x,y\rangle=0$ then $\langle T(x),T(y)\rangle=0$.

*Show that columns of the matrix of $T$ in the standard basis of $\mathbb{R}^n$ are mutually orthogonal. 


For the answer
I have attempted using the basic definition of inner product with the norm.
That is $||v||^2=<v,v>$. Also with the Cauchy Schwarz inequality. But couldn't make any progress. 

If
$\Vert T(z) \Vert = \Vert z \Vert, \; \forall z \in \Bbb R^n, \tag 1$
then
$\langle T(z), T(z) \rangle = \Vert T(z) \Vert^2 = \Vert z \Vert^2 = \langle z, z \rangle, \; \forall z \in \Bbb R^n, \tag 2$
so with
$z = x + y, \tag 3$
$\langle T(x + y), T(x + y) \rangle = \langle x + y, x + y \rangle; \tag 4$
now,
$\langle T(x + y), T(x + y) \rangle = \langle T(x) + T(y), T(x) + T(y) \rangle$
$= \langle T(x), T(x) \rangle + \langle T(x), T(y) \rangle + \langle T(y), T(x) \rangle + \langle T(y), T(y) \rangle, \tag 5$
and
$\langle x + y, x + y \rangle = \langle x, x, \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle; \tag 6$
we combine (4)-(6) and find
$\langle T(x), T(x) \rangle + \langle T(x), T(y) \rangle + \langle T(y), T(x) \rangle + \langle T(y), T(y) \rangle$
$= \langle x, x, \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle, \tag 7$
and using (2) with $z = x, y$ we write
$\langle T(x), T(y) \rangle + \langle T(y), T(x) \rangle = \langle x, y \rangle + \langle y, x \rangle, \tag 8$
and since for any $u, v \in \Bbb R^n$
$\langle u, v \rangle = \langle v, u \rangle, \tag 9$
we find that (8) yields
$2\langle T(x), T(y) \rangle = 2 \langle x, y \rangle, \tag{10}$
whence
$\langle T(x), T(y) \rangle = \langle x, y \rangle, \; \forall x, y \in \Bbb R^n, \tag{11}$
it now readily follows that
$\langle x, y \rangle = 0 \Longleftrightarrow \langle T(x), T(y) \rangle = 0; \tag{12}$
returning to (11), we also have
$\langle x, y \rangle = \langle T(x), T(y) \rangle = \langle x, T^TT(y) \rangle = \langle T^TT(x), y \rangle, \; \forall x, y \in \Bbb R^n, \tag{13}$
and thus we conclude that
$T^TT = TT^T = I; \tag{14}$
now if we write $T$ in columnar form
$T = \begin{bmatrix} \mathbf t_1 & \mathbf t_2 & \ldots & \mathbf t_n \end{bmatrix}, \tag{15}$
i.e., the vectors $\mathbf t_i$, $1 \le i \le n$, are the columns of $T$, then
$T^T = \begin{bmatrix} \mathbf t_1^T \\ \mathbf t_2^T \\ \vdots \\ \mathbf t_n^T \end{bmatrix}, \tag{16}$
that is, the rows of $T^T$ are the $\mathbf t_i^T$, then it follows from (14) that
$\begin{bmatrix} \mathbf t_i^T \cdot \mathbf t_j \end{bmatrix} = \begin{bmatrix} \mathbf t_1^T \\ \mathbf t_2^T \\ \vdots \\ \mathbf t_n^T \end{bmatrix} \begin{bmatrix} \mathbf t_1 & \mathbf t_2 & \ldots & \mathbf t_n \end{bmatrix} = T^TT = I, \tag{17}$
from which it is seen that
$ \mathbf t_i^T \cdot \mathbf t_j = \delta_{ij}, \; 1 \le i, j \le n, \tag{18}$
that is, the $\mathbf t_i$, the columns of $T$, are orthonormal vectors.  $OE\Delta$.
