# Inclusion between diagonalizable , unitary, normal and hermitian?

What is the inclusion between the following subsets of $$\mathbb{C}^{n \times n}$$: diagonalizable, unitary, normal and hermitian.

I will call the sets $$\mathcal{D}$$, $$\mathcal{U}$$, $$\mathcal{N}$$ and $$\mathcal{H}$$.

$$\mathcal{H}\subset\mathcal{N}$$ because $$AA^*=A^2=A^*A$$, remember that $$A^*=A$$.

$$\mathcal{U}\subset\mathcal{N}$$ because $$AA^*=I=A^*A$$, remember that $$A^*=A^{-1}$$.

$$\mathcal{N}\subset\mathcal{D}$$: It is a well-known theorem (for example you can find in Hoffman's book Linear Algebra) that every normal matrix over $$\mathbb{C}$$ is (unitarily) diagonalizable.

$$\mathcal{D}$$ is not contained in any of the other sets, because for example $$\begin{pmatrix} 1&2\\0&2 \end{pmatrix}$$ is diagonalizable (it has two distinct eigenvalues, namely 1 and 2) but it is not normal because $$AA^T=\begin{pmatrix} 5&4\\4&4\end{pmatrix}$$ and $$A^T A=\begin{pmatrix} 1&2\\2&8\end{pmatrix}$$. This also implies that $$\mathcal{D}\not\subset \mathcal{H}, \mathcal{U}$$ either.

$$\mathcal{H}\not\subset \mathcal{U}$$ because any real diagonal matrix is Hermitian (as $$A^*=A^T=A$$) but it does not necessarily have $$|\det A|=1$$. This also implies that $$\mathcal{N}\not\subset \mathcal{U}$$.

$$\mathcal{U}\not\subset \mathcal{H}$$ because for example $$\begin{pmatrix} 0&-1\\1&0 \end{pmatrix}$$ is unitary ($$AA^*=AA^T=I$$) but is not Hermitian because $$A^*=A^T\neq A$$. This also implies that $$\mathcal{N}\not\subset \mathcal{H}$$.

• Thank toy for the concise and clear answer! Commented Aug 27, 2020 at 18:02