Is the sum over the fourth powers of matrix elements basis dependent?

Question

Given a matrix $A=(A_{ab})_{a,b=1}^n$ the quantity $$\sum_{a,b} \lvert A_{ab}\rvert^2$$ appears to depend on the chosen basis. But it turns out that $$\sum_{a,b} \lvert A_{ab}\rvert^2=\sum_a (AA^\ast)_{aa}=\mathbf{tr}AA^\ast,$$ where $A^\ast$ is the adjoint of $A$. In particular, $\sum_{a,b}\lvert A_{ab}\rvert^2$ does not depend on the chosen basis as the trace is basis independent.

What about $\sum_{a,b} \lvert A_{ab}\rvert^4$? Does this depend on the chosen basis? If not, what is a basis-free way to write it?

Answer 1

Your claim about sum of squared entries is wrong. The sum actually depends on the basis. It's just invariant under a change of orthonormal basis. In other words, if $B=P^{-1}AP$ for some matrix $P$ that not orthogonal, it can happen that $\sum_{i,j}|a_{ij}|^2\ne\sum_{i,j}|b_{ij}|^2$.

For the sum of fourth powers, the answer is no, even if we perform a change of basis orthogonally. It is easy to obtain such an example: $$ A=\pmatrix{1&2\\ 3&4},\ B=\left[\frac1{\sqrt{2}}\pmatrix{1&-1\\ 1&1}\right]^{-1}A\left[\frac1{\sqrt{2}}\pmatrix{1&-1\\ 1&1}\right]=\pmatrix{5&1\\ 2&0}. $$ The sum of fourth powers of all entries in $A$ is 354 and its counterpart in $B$ is 642.

Answer 2

Consider $A:=\begin{pmatrix} 1 & 1\\0 &1\end{pmatrix}$. The magic number is $3$.

Now look what happens when we change the basis to $(0,1)^T$, $(1,1)^T$. A quick calculation shows that the matrix becomes $B:=\begin{pmatrix} 0 & -1\\1 &2\end{pmatrix}$ whose magic number is $18$.