Assume I exchange two rows of a square complex $n\times n $ matrix.
Are the Euclidean norm and the Hilbert-Schmidt norm of the new matrix (obtained from the first one by exchanging two of its rows) the same as the orginal one?
Assume I exchange two rows of a square complex $n\times n $ matrix.
Are the Euclidean norm and the Hilbert-Schmidt norm of the new matrix (obtained from the first one by exchanging two of its rows) the same as the orginal one?
If by the Euclidean norm you mean the operator norm induced by the Euclidean norm on vectors, $\lVert A\rVert_2 = \max_{x\ne 0} \frac{\lVert Ax\rVert_2}{\lVert x\rVert_2},$ then this follows easily from the facts that permuting the rows of $A$ is the same as premultiplying it by an appropriate permutation matrix $P$, and that $\lVert Py \rVert_2 = \lVert y\rVert_2$ for any vector $y$.
Hilber-Schmidt norm is defined by
$\|A\|_{HS}^2:=\sum_{i,j}|A_{i,j}|^2$
Check the wikipedia page.. Hence we see that this sum is independent of interchange the rows or column. Also Euclidean norm sees $A$ as point in $C^n$ and hence norm is independent if we exchange the rows.
These are both entrywise norms, and so...