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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?

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  • $\begingroup$ It sometimes helps if you take a look at your posting. What you had between each $\langle$ and $\rangle$ was invisible. Putting $\$$ around the formulas (to make them into TeX) made them reappear. $\endgroup$ – Robert Israel Nov 12 '13 at 0:21
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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}

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  • $\begingroup$ Hmmmm, that's pretty clever. Did you mean for the last line to be ||UA|| though? Could you please elaborate a bit more on the conjugate transpose stuff a bit? Thanks for the help! $\endgroup$ – Incognito Nov 11 '13 at 23:55
  • $\begingroup$ @user108149 Now I think my answer clearer for you enteder. I have helped. $\endgroup$ – MathOverview Nov 12 '13 at 12:08
  • $\begingroup$ Thanks for the super clear response; that was very helpful. $\endgroup$ – Incognito Nov 13 '13 at 2:54
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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 $(A, B) = {\rm trace}(A^* B)$.

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  • $\begingroup$ Well, usually, if a norm is not induced from an inner product space, you can't necessarily assume that the inner product exists. Hence, I wasn't sure if I could do what i was doing up there with inner products. Though, it sounds like any frobenius norm automatically has an inner product, based upon your response. Thank you! $\endgroup$ – Incognito Nov 11 '13 at 23:50
  • $\begingroup$ The notation is suggestive, but there's a subtle point here. Multiplication on the left by matrix $T$ is an operator on this inner product space. You want to show that the adjoint of that operator is multiplication on the left by $T^*$ (the adjoint of the matrix), i.e. that $(T A, B) = (A, T^*B)$. It's true, and easy, but does require stating. $\endgroup$ – Robert Israel Nov 12 '13 at 0:31

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