Normal but not hermitian nor unitary

linear-algebra matrices

Solution 1:

The answer to your (imprecise) question lies in the Spectral theorem for normal matrices: normal matrices are precisely those that are unitarily diagonalizable. Hermitian and unitary matrices are special cases: hermitian matrices are normal with real eigenvalues, while unitary matrices are normal with complex eigenvalues of modulus one.

Therefore to answer your question, you should look for some matrix with complex non-real and non-unitary eigenvalues.

Solution 2:

Hope this matrix can work

$$A = \begin{pmatrix} 1 & 1+i & 1 \\ -1+i & 1 & 1 \\ -1 & -1 & 1\end{pmatrix}$$

$AA' = A'A$ so normal.

$AA' \neq 0$ so not unitary.

$A \neq \bar{A}'$ so not hermitian.

Related

Conditional return time of simple random walk
Intuition of joint density of min(X,Y) and max(X,Y)
Exercise 4.9, Chapter I, in Hartshorne
$(e_{n})$ orthonormal basis, $(f_{n})$ orthonormal sequence such that $\sum\left\|e_{n}-f_{n}\right\|^{2}<\infty$ Then $(f_{n})$ is orthonormal basis.
Separable ODE and singular solutions
Why is $ U \otimes \operatorname{Ind}(W) = \operatorname{Ind}(\operatorname{Res}(U) \otimes W)$?
Find the principal value of $\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx$
Given $2n$ points in the plane, $n$ blue points and $n$ red points, no $3$ are collinear. Prove that we have at least two "balanced lines".
Orthogonal projection matrix
How to interpret Lagrangian function (specifically not Lagrangian multiplier)
Generators of the ideal correspond to $d$-uple embedding
If $f'$ is bounded, then $f$ is uniformly continuous