Properties of length preserving linear transformations

Question

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:

• 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.

Answer 1

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. $OE\Delta$.

Answer 2

Do you understand what it means for a set of vectors to be mutually orthogonal?

In case you don't, it's just saying that for all vectors in the set

$v_i\cdot v_j = 0$ for all $i \neq j$

Consider then the definition of a linear transformation that preserves length, there are two other important parts other than $<x, y> = <T(x), T(y)>$

How can you use the definition there and the definition of an orthogonal matrix to show the orthogonality of $T$?

Answer 3

The first claim follows readily from the polarization identity: $\langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|)$.

For the second claim, observe that the i-$th$ column of the matrix for $T$ is given by the coefficients on the right side of the equation $Te_i=\sum_ka_{ki}e_k.$ The result now follows by the first claim since

$i\neq j\Rightarrow \langle e_i,e_j\rangle=0\Rightarrow \langle Te_i,Te_j\rangle=$

$\langle\sum_ka_{ki}e_k,\sum_ka_{kj}e_k \rangle =\sum_ka_{ki}a_{kj}=0,$

as desired.