Let $A^*$ denote the complex conjugate transpose of a matrix $A$, and $\|\cdot\|$ the norm induced by the Euclidean norm (2-norm). Define

$$k(A)=\frac{1}{2}\|A-A^*\|$$ and $$m(A)=\max_{\lambda\in\sigma(A)}| \text{Im}\:\lambda |.$$ where $\sigma(A)$ is the spectrum. Note that

  1. If $A$ is Hermitian then $k(A)=m(A)=0$.
  2. If $A$ is skew-Hermitian then $k(A)=m(A)$.

Now, assume $A$ is normal. Can we find any relations? Upper or lower bounds connecting $m$ and $k$?


If $A$ is normal, $k(A) = m(A)$. This is easily seen by unitarily diagonalizing $A$.

  • $\begingroup$ Unitarily diagonalizing $A$ only leads us to $k(A)=\frac{1}{2}||A-A^*||=\frac{1}{2}||UAU^*-U^*A^*U||=\frac{1}{2}||AU^*-U^*A^* ||$, and I don't know how to relate this to the imaginary parts of the eigenvalues. $\endgroup$ – PeterA Feb 2 '15 at 8:44
  • $\begingroup$ @PeterA Well, if you take $A=UDU^*$, then $k(A)=\frac{1}{2}\|D-\bar{D}\|$. If $\lambda=\alpha+i\beta$ with real $\alpha$ and $\beta$ is a diagonal entry of $D$ (eigenvalue of $A$), the corresponding entry of $D-\bar{D}$ is $\lambda-\bar{\lambda}=(\alpha+i\beta)-(\alpha-i\beta)=2i\beta$. Taking a norm of $\|D-\bar{D}\|$ gives hence exactly $2m(A)$. $\endgroup$ – Algebraic Pavel Feb 2 '15 at 9:25
  • $\begingroup$ @AlgebraicPavel Okay, so if $A=UDU^*$, then $A^*=U^*\bar{D}U$ and $A-A^*=UDU^*-U^*\bar{D}U$. The norm is then $$\|A-A^*\|=\|UDU^*-U^*\bar{D}U\|=\|DU^*-U^*\bar{D}\|$$ But how to get from here to $\|D-\bar{D}\|$? $\endgroup$ – PeterA Feb 2 '15 at 9:33
  • $\begingroup$ @PeterA $\|A-A^*\|=\|UDU^*-U\bar{D}U^*\|=\|D-\bar{D}\|$. $\endgroup$ – Algebraic Pavel Feb 2 '15 at 9:49
  • $\begingroup$ @AlgebraicPavel So, you mean $A=UDU^*$ then $A^*=U\bar{D}U^*$, and also $A^*=U^*\bar{D}U$. But then $U=U^*=U^{-1}$? I guess I have to refresh my memory. $\endgroup$ – PeterA Feb 2 '15 at 9:55

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