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I'm considering the norm defined on matrices by

$$|A|_F = sqrt{sum_{i,j}|a_{ij}|^2}$$

I want to show that it is unitarily invariant, so that for unitary $U$ we have that

$$|UA|_F = |A|_F = |AU|_F$$

however I have trouble doing it directly. Writing $|UA|_F$ directly I find by Cauchy-Schwarz that

$$|UA|_F = sqrt{sum_{i,j}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2}= sqrt{sum_{i,j}|langle U_i,overline{A_j}rangle|^2}leq sqrt{sum_{i,j}|A_j|^2}$$

where $U_i$ denotes the $i$th row of $U$ and $A_j$ the $j$th column of $A$. However this estimate is to crude and will not equal $|A|_F$. I would like to prove this without refering to trace or singular values and would appreciate a hint, rather than a full solution, on how to tackle this problem.

EDIT: Completion of the proof based on the answer from $A.Gamma$:

Since the rows of $U$ constitute an orthonormal basis for $mathbb{C}^n$ we find by Parsevals theorem that

$$|UA_j|_2^2 = sum_{i=1}^{n}left|sum_{k=1}^{n}u_{ik}a_{kj}right|^2 = sum_{i=1}^{n}|langle U_i,overline{A_j}rangle|^2 = |overline{A_j}|_2^2 = |A_j|_2^2$$

Since
$$
UA=[UA_1 UA_2 ldots UA_n]
$$
you need to prove that
$$
|UA|_F^2=sum_{j=1}^n|UA_j|_2^2stackrel{?}{=}sum_{j=1}^n|A_j|_2^2=|A|_F^2.
$$
It suffice to prove that $|UA_j|_2^2=|A_j|_2^2$.

P.S. For $AU$ use conjugation.

Quick and dirty:

$$|UA|_F^2 = operatorname{Tr}((UA)^*(UA)) = operatorname{Tr}(A^*U^*UA) = operatorname{Tr}(A^*A) = |A|_F^2$$
and then since $U^*$ is also unitary
$$|AU|_F^2 = |(U^*A^*)^*|_F^2 = |U^*A^*|_F^2 = |A^*|_F^2 = |A|_F^2$$

Alternative argument:

Note that $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_n}$ for $mathbb{C}^n$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.

We have
$$sum_{i=1}^n |Au_i|_2^2 = sum_{i=1}^n langle Au_i, Au_irangle = sum_{i=1}^n langle A^*Au_i, u_irangle = sum_{i=1}^n lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$

because the trace is the sum of eigenvalues.

The interesting part is that the sum $sum_{i=1}^n |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_n}$. Indeed, if ${v_1, ldots, v_n}$ is some other orthonormal basis for $mathbb{C}^n$, we have
begin{align}
sum_{i=1}^n |Au_i|_2^2 &= sum_{i=1}^n langle A^*Au_i, u_irangle\
&= sum_{i=1}^n leftlangle sum_{j=1}^nlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^nlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^n sum_{k=1}^n left(sum_{i=1}^nlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^n sum_{k=1}^n langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^n langle A^*A v_j,v_jrangle\
&= sum_{j=1}^n |Av_j|_2^2
end{align}

Now, if $U$ is unitary, for any orthonormal basis ${u_1, ldots, u_n}$ we have that ${Uu_1, ldots, Uu_n}$ is also an orthonormal basis so:

$$|AU|_F^2 = sum_{i=1}^n |A(Ue_i)|^2 = |A|_F^2$$