Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 0 down vote accepted

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 $y$.

share|improve this answer

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.

share|improve this answer

These are both entrywise norms, and so...

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.