Does showing that a matrix over a complex inner product space is an isometry mean that it is a unitary matrix?

Question

I'm given the problem of showing that the change of basis matrix between two orthonormal bases of a complex inner product space is a unitary matrix. A solution to this problem is given here: prove change of basis matrix is unitary

My book gives the definition that an $n \times n$ complex matrix is said to be unitary if the conjugate transpose of the matrix is equal to the inverse of the matrix. I also have a theorem that says $T$, a linear operator on an inner product space $V$, is an isometry if and only if the inner product of $T(v)$ against $T(w)$ is equal to the inner product of $v$ against $w$ for all $v,w \in V$. In the given proof he has the change of basis matrix given by M, and says we need to show that the inner product of $Mx$ against $My$ is equal to the inner product of $x$ against $y$ for all $x,y \in V$ in order to show that $M$ is a unitary matrix. But nowhere does it have that the conjugate transpose of the matrix is equal to its inverse. Is saying that a complex matrix is unitary basically just saying that the linear operator represented by the matrix is an isometry? Any insight would be much appreciated.

Answer 1

Yes. If $A \in M_n(\mathbb{C})$ and $T_A \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ is the corresponding linear operator $T_A(v) = Av$ then $A$ is unitary (in the sense that $AA^{*} = A^{*}A = I$) if and only if $T_A \colon (\mathbb{C}^n, \left< \cdot, \cdot \right> \rightarrow (\mathbb{C}^n, \left< \cdot, \cdot \right>)$ is a unitary map in the sense of inner product spaces where $\left< \cdot, \cdot \right>$ is the standard inner product (we have $TT^{*} = T^{*}T = \mathrm{id}|_{\mathbb{C}^n}$ where $T^{*}$ is the adjoint operator). This is equivalent to $T_A \colon (\mathbb{C}^n, \left< \cdot, \cdot \right> \rightarrow (\mathbb{C}^n, \left< \cdot, \cdot \right>)$ being an isometry (in the sense that $\left< T_A(x), T_A(y) \right> = \left< x, y \right>$ for all $x,y \in \mathbb{C}^n$).

More generally, if $(V, \left< \cdot, \cdot \right>)$ is an inner product space and $\mathcal{B}$ is an orthonormal basis then the adjoint of $T^{*}$ is represented in the basis $\mathcal{B}$ by the adjoint matrix of $T$. That is, $[T^{*}]_{\mathcal{B}} = ([T]_{\mathcal{B}})^{*}$. This generalizes the previous case as the standard basis for $\mathbb{C}^n$ is an orthonormal basis.