I have read different definitions, or properties, of a Hermitian matrix, and still am not sure if I have a sufficient number of properties to define a Hermitian matrix.
Suppose the following is true about Hermitian Matrices.
$M\vec{a} = \lambda\vec{a}$
where all scalars $\lambda$s (eigenvalues) are real.
Lets just use a 2D matrix.
If $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $det\begin{bmatrix} a-\lambda & b \\ c & d-\lambda \end{bmatrix}=0$.
This means that (by quadratic solution to characteristic equation) $$\lambda = \frac{(a+d)\pm\sqrt{(a+d)^2-4*(a^2-cb)}}{2}$$
For lambda to be real (in both + and - cases), there are certain restrictions that must be obeyed. However they are not as restrictive as the definition of a Hermitian matrix.
One restriction is that $$Im(a+d)=0$$ (imaginary components cancel). Another is $$(a+d)^2-4*(a^2-cb) >= 0 $$ (>= or just >?) and simultaneously $$Im((a+d)^2-4*(a^2-cb))=0$$
Even though Hermitian matrices do not allow imaginary diagonal elements, I'm pretty sure I could think of a matrix that would satisfy these conditions with imaginary $a$ and $d$ elements.
I'm guessing I'm not understanding the properties of a Hermitian matrix. What am I missing?
By definition, a complex square matrix $A$ is hermitian if and only if $A^H = A$. Where $A^H = \bar A^T$ denotes the conjugate transposed of $A$.
It is the result of the spectral theorem, that all eigenvalues of a hermitian matrix are real and there exists eigenvector basis.
The matrix $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$ has clearly only real eigenvalues, but is not hermitian.