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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?

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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 $y$.

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Hilber-Schmidt norm is defined by


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.

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These are both entrywise norms, and so...

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I thought (and Wikipedia agrees) that the Euclidean norm of a matrix was the operator norm $\lVert A \rVert_2 = \max_{x\ne0}\lVert Ax\rVert_2/\lVert x\rVert_2$, and that's not an entrywise norm. If Wikipedia is incorrect, someone should correct it. If the term is ambiguous, the asker should clarify which meaning is intended. – Rahul Mar 1 '12 at 3:04
I guess it is ambiguous! I was using this – Juan S Mar 1 '12 at 4:29

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