Frobenius Norm Invariant to Orthonormal Transform Proof

We want to proof:

∣∣AU∣∣F=∣∣UA∣∣F=∣∣A∣∣F||AU||_F=||UA||_F = ||A||_F

The definition of Frobenium Norm:

∣∣A∣∣F=∑i,jAij2=trace(A⊤A)||A||_F = \sqrt{\sum_{i,j} A_{ij}^2} = \sqrt{trace(A^{\top}A)}

Nice property about trace:

trace(AB)=trace(BA)trace(AB)=trace(BA)

Note:

∣∣AU∣∣F=trace((AU)⊤AU)=tr(U⊤A⊤AU)=tr(UU⊤A⊤A)=∣∣A∣∣F\begin{split} ||AU||_F &= \sqrt{trace((AU)^{\top}AU)} \\ &= \sqrt{tr(U^{\top}A^{\top}AU)} \\ &= \sqrt{tr(UU^{\top}A^{\top}A)} \\ &= ||A||_F \end{split}

Same way the other side…