# Normal but not hermitian nor unitary

I have to find out a normal transformation that is neither hermitian nor unitary. http://en.wikipedia.org/wiki/Normal_matrix gives me the answer. However, I would like to know how to find it out mathematically, not just guess and test. Can I go from the properties of eigenvalues of hermitian and unitary to get the answer? For example: eigenvalues of a hemitian must be real, then I choose (i,-i,0) as eigenvalues of the required matrix. Those eigenvalues satisfy the condition that the required matrix is not unitary whose eigenvalues are |1|. From those, I have characteristic equation has the form of x(x^2+1)= 0 which leads to the form of the required matrix has the trace =0 and det = 0. Can I do it?

• Hmm, to find out what? – user1551 Apr 27 '14 at 14:19
• find out the normal matrix that is neither hermitian nor unitary – user73195 Apr 27 '14 at 14:21

## 2 Answers

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.

• Thanks for the answer. You mean I can arbitrary choose a non-real and non-unitary eigenvalues and then go back ward to construct a normal matrix, right? for example, if I pick (i, -i)as eigenvalues of A, then A will have tr(A) =0 and det(A) =1. And construct A based on those conditions, am I right? – user73195 Apr 27 '14 at 14:50
• Eigenvalues $\pm i$ won't work as they will give rise to a unitary matrix. But apart from that, that's ok. More precisely, you can take any finite sequence $(z_1\ldots z_N)$ of complex numbers and a unitary matrix $U$ to obtain a normal matrix by conjugation: $N=U^\star\mathrm{diag}(z_1\ldots z_N) U$. The content of the spectral theorem is that any normal matrix is obtained that way. – Giuseppe Negro Apr 27 '14 at 15:35

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.

• Thanks for the answer. However, what I want to know is the method to find it out without guessing. I have A = {{i,o},{0,2}} satisfies the condition, too. All I can do is writing some normal matrix and then testing. – user73195 Apr 27 '14 at 14:41
• Only trial and error method as it is easiest here. Otherwise you may find an operator over some complex vector space which is normal but not unitary and harmitian. Then take its matrix representation. The matrix will also suit. – Supriyo Apr 27 '14 at 14:44
• BTW, if A^2 =A (idempotent), does it imply (A^*)^2 = A^*? – user73195 Apr 27 '14 at 15:00