Frobenius Norm with Unitary Operators For something I'm working on, I have a matrix $A$ with another matrix $U$ which is unitary ($U^*U = I$), and I'm trying to show that, for the Frobenius norm, $\|A\| =\|UA\|$.  Now, I can do this pretty easily if an inner product space exists.  For example, $\|A\| = \sqrt{\langle A,A\rangle}$ and $\|UA\| = \sqrt{\langle UA,UA\rangle} = \sqrt{\langle A,U^*UA\rangle} = \sqrt{\langle A,A\rangle} = \|A\|$.  However, I'm not sure if I evoke the inner product space if I'm just told that the Frobenius norm exists.  Is this the appropriate approach?
 A: By definition of Frobenius norm of a matrix $A\in\mathbb{C}^{n\times n}$, 
$$
\| A\|_F:=\sqrt{\mbox{trace}(A^*A)}=\ldots =\sqrt{\sum_{j=i}^n\sum_{i=1}^n\overline{A_{ji}}\cdot A_{ij}}
$$
This norm is a norm defined by an inner product $\langle \cdot , \cdot \rangle$ (i.e. $\|A\|_F:=\sqrt[2]{\langle A ,  A\rangle}$ ). In this case, for $A=(A_{\,\alpha\,\beta})_{n\times n}\in\mathbb{C}^{n\times n}$ and $B=(B_{\,u\,v})_{n\times n}\in\mathbb{C}^{n\times n}$ we have
\begin{align}
\langle A, B\rangle_F := & \mbox{trace}\bigg(A^* B\bigg)\\
\end{align}
Remenber that, for all $X=(X_{\,i\,j} )_{n\times n}\in\mathbb{C}^{n\times n}$, $\mbox{trace}(X)=\sum_{\ell=1}^{n}X_{\ell\ell}$ and 
\begin{align}
C^*\cdot D=&(C_{ij})_{n\times n}^*\cdot(D_{uv})_{n\times n} \\
          =&(\overline{C_{ji}})_{n\times n}\cdot(D_{uv})_{n\times n} \qquad \mbox{ definition of $*$ } \\
        =& \bigg(\sum_{k=1}^{n}C_{jk}\cdot D_{kv} \bigg)_{n\times n}\qquad \mbox{ rule of matrix product } 
\end{align}
Now, your question
\begin{align}
\| A\|:=&\sqrt{\mbox{trace}(A^*A)}\\
       =&\sqrt{\mbox{trace}(A^*IA)}\\
       = & \sqrt{\mbox{trace}(A^*[U^*U]A)}\\
       = & \sqrt{\mbox{trace}([A^*U^*][UA])}\\
       = & \sqrt{\mbox{trace}([UA]^*[UA])}\\
=& \| UA\| 
\end{align}
A: I don't know what you mean by "if an inner product space exists".  The Frobenius norm does come from an inner product, namely the Frobenius inner product
$(A, B) = {\rm trace}(A^* B)$.  
