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This considers any operator norm. Given a matrix $\bf A$, Let $\lambda({\bf A})$ be its spectrum, and ${\bf J}_\lambda$ be the Jordan block corresponding to eigenvalue $\lambda$. Do we have $\|{\bf A}\|=\max_{\lambda \in \lambda({\bf A})}\|{\bf J}_\lambda\|$ for any operator norm?

For 2-norm, it is true, see a previous post Is the matrix norm of a matrix equal to the maximum of the norms of its Jordan block?.

If the claim does not hold for any operator norm, does it hold for some matrix norm other than 2-norm?

Thanks!

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  • $\begingroup$ Don't think it's true for Frobenius norm. Or do you only care about induced norms? (Even so I also doubt it's true for $1$-norm, for the same reason that it's suspicious for the F-norm: Jordan blocks can be of much smaller sizes and much sparser.) $\endgroup$
    – Vim
    Commented Sep 30, 2017 at 16:29
  • $\begingroup$ @Vim Thanks for comment, but Frobenius norm is not an operator norm. Yes, I also think it is not true for 1-norm or infinite norm. But I hope someone could give a more accurate characterization of the claim. $\endgroup$
    – Tony
    Commented Sep 30, 2017 at 16:39

1 Answer 1

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This is false even for the $2$-norm, which I am guessing is the operator norm for operators on $\mathbb{R}^n$ with its usual inner product. Consider for instance the operator on $\mathbb{R}^2$ with the usual inner product, which, written with respect to an orthonormal basis, has the matrix

$$ \begin{bmatrix} 1 & 1000\\ 0 & 2 \end{bmatrix}. $$

The eigenvalues of this matrix are $1$ and $2$, but its operator norm is at least $1000$. The reason why the norm is not $2$ is that the decomposition into eigenspaces is not orthogonal.

Some orthogonality assumption is implicit in the answer that you cite, which is assuming (at least) that $\mathbb{R}^n$ is equipped with an inner product that makes the generalized eigenspaces orthogonal.

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