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I am trying to proof the following inequality (source, equation 17):

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where $C \in R^{n\times n}$. I tried the following:

$$||C (\Lambda - \Lambda^T)||_F \le ||C||_F ||\Lambda - \Lambda^T || = \sqrt{\sum \lambda_i(C)} || \Lambda - \Lambda^T||_F$$

But I don't understand how to pursue farther. How to prove that the square root of the eigenvalues summation is less than the max eigenvalue?

rando
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  • is $C$ normal? If not the statement is false -- e.g. select any non-zero nilpotent $C$. On the other hand, if $C$ is normal, there are proofs here: https://math.stackexchange.com/questions/3538652/frobenius-norm-and-operator-norm-inequality – user8675309 Aug 15 '20 at 17:59

1 Answers1

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The $C$ in the paper is in the form of $(A^TA)^{-1}$ (eq. 7). So, it is positive definite and by a change of orthonormal basis, we may assume that $C$ is a nonnegative diagonal matrix. The inequality in question is now straightforward if you directly expand both sides entrywise. Alternatively, for any matrix $X=\pmatrix{x_1&x_2&\ldots&x_m}$, we have $$ \|CX\|_F^2=\sum_j\|Cx_j\|_2^2\le\sum_j\|C\|_2^2\|x_j\|_2^2=\lambda_\max(C)^2\|X\|_F^2. $$ Since $C$ is positive definite, if we take square roots on both sides, the result follows.

user1551
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  • didn't get why $||C||2^2 = \lambda{max}(C)^2$. Can you elaborate? – rando Aug 15 '20 at 18:50
  • @randomndrandom $C$ is positive definite. Therefore its largest eigenvalue is equal to its largest singular value. – user1551 Aug 15 '20 at 19:30
  • No problem with that. I meant $||C||_2^2 = \sum \lambda_i(C)$ how did it end up equals to the max eigenvalue? – rando Aug 17 '20 at 15:51
  • @randomndrandom You seem to have misunderstood the meaning of $|C|_2$. It is the standard notation for the induced 2-norm of $C$, i.e. the largest singular value of $C$. – user1551 Aug 17 '20 at 17:59