# The matrix of an isometry has orthonormal columns

Axler's Linear Algebra Done Right proves that if $$T : V \to V$$ is a linear operator on a finite-dimensional inner product space over $$F \in \{ \mathbb{R}, \mathbb{C} \}$$, then the following are equivalent to $$T$$ being an isometry.

1. $$||T v|| = ||v||$$ for all $$v \in V$$.
2. $$\langle T u, Tv \rangle = \langle u, v \rangle$$ for all $$u, v \in V$$.
3. $$T e_1, \dots, T e_r$$ is orthonormal for any orthonormal list of vectors $$e_1, \dots, e_r$$ in $$V$$.
4. There exists an orthonormal basis $$e_1, \dots, e_n$$ of $$V$$ such that $$T e_1, \dots, T e_n$$ is orthonormal.
5. $$T^* T = I$$.
6. $$T T^* = I$$.
7. $$T$$ is invertible with inverse $$T^{-1} = T^*$$.
8. $$||T^* v|| = ||v||$$ for all $$v \in V$$.

I proved all of the equivalences except (1.) $$\implies$$ (2.) without consulting the book's proof, but I am stumped on how to justify the following comment in the book regarding this result.

The equivalence of (1.) and (3.) [or (4.)] shows that "an operator is an isometry if and only if the list of columns of its matrix with respect to every [or some] orthonormal basis is orthonormal." Why is this true?

Note that the text accidentally omits the first occurrence of "orthonormal" in the statement above (see https://linear.axler.net/LADRErrataThird.html).

I'm confused because the columns of the matrix are not simply $$T$$ applied to each basis element, but the coefficients of those vectors with respect to the basis. And it's not clear what inner product is being used on the columns of the matrix. I assume it's the Euclidean inner product $$\langle (w_1, \dots, w_n), (z_1, \dots, z_n) \rangle := w_1 \overline{z_1} + \dots + w_n \overline{z_n}$$.

Suppose that $$T$$ is an isometry. Then by (4) there is an orthonormal basis $$e_1,\dots,e_n$$ of $$V$$ such that $$Te_1,\dots,Te_n$$ is also an orthonormal basis. Theorem 6.30 gives us that $$(\forall j \in \{1,\dots,n\}) \quad Te_j = \langle Te_j,e_1 \rangle e_1 + \cdots + \langle Te_j,e_n \rangle e_n$$ so, the $$j$$-th column of the associated matrix is $$a_j := \begin{pmatrix} \langle Te_j,e_1 \rangle \\ \vdots \\ \langle Te_j,e_n \rangle \end{pmatrix} \in F^{n \times 1}.$$ Then, as you say, if we consider the Euclidean inner product on $$F^{n \times 1} \cong F^n$$ we have \begin{align} \langle a_j, a_k \rangle &= \sum_{i=1}^n \langle Te_j,e_i \rangle \overline{\langle Te_k,e_i \rangle} \\ &= \sum_{i=1}^n \langle Te_j, \langle Te_k,e_i \rangle e_i \rangle \qquad (\langle -,\lambda v \rangle = \overline \lambda \langle -,v \rangle) \\ &= \left\langle Te_j, \sum_{i=1}^n \langle Te_k,e_i \rangle e_i \right\rangle = \langle Te_j, Te_k \rangle \end{align} so $$a_1,\dots,a_n$$ is indeed an orthonormal list.
Conversely, if there is an orthonormal basis $$e_1,\dots,e_n$$ of $$V$$ such that the associated matrix $$A$$ of $$T$$ has orthonormal columns, then $$A^*A = I$$ (if $$B$$ and $$C$$ are square matrices, the $$(i,j)$$-th entry of $$B^*C$$ is precisely the Euclidean inner product of the $$j$$-th column of $$C$$ with the the $$i$$-th column of $$B$$) and theorem 7.10 now gives us $$T^*T=I$$, so $$T$$ is an isometry by (5).