For an $n \times n$ square complex matrix let say $A$ with eigenvalues $\lambda_1,\lambda_2,.....,\lambda_n$. $A$ is normal iff $A$ is unitary diagonalizable;that is there exist unitary matrix U such that

$U*A*U=\text{diag}(\lambda_1, \lambda_2,...,\lambda_n)$.

Above is the definition for spectral decomposition. Now my question is what are the differences in the spectral decomposition of normal,hermitian ,+ve semidefinite and unitary matrices?


When $A$ and $B$ are normal, since spectral radii and operator norms coincide, we always have $\rho(A-B)=\|A-B\|\le\|A\|+\|B\|=\rho(A)+\rho(B)$ and this bound is sharp.

Other than that, I don't think one can say much because $A-B$ can be quite arbitrary. For instance, any complex square matrix with unit Frobenius norm is the average of two unitary matrices (cf. Li and Poon , "Additive Decomposition of Real Matrices", Lin. Mult. Alg. 50(4): 321–326, 2002.). So, $\{A-B: A,\,B\in U(n)\}$ at least contains a very rich bounded subset of $M_n(\mathbb C)$.

However, when $A$ and $B$ are Hermitian, Weyl's inequality applies.


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