Proof that if a matrix has real eigenvalues and orthogonal eigenvectors then it's Hermitian I saw many times that a Hermitian matrix has real eigenvalues and orthogonal eigenvectors and I pretty much know the proof of that.
What I was wondering is the other way around, if a matrix has real eigenvalues and orthogonal eigenvectors, does it make it Hermitian?
(I use Dirac's bra/ket notation)
Let $\left|v_1\right>$ and $\left|v_2\right>$ be two distinct orthogonal eigenvectors with distinct eigenvalues $\lambda$ and $\mu$ respectively for $A$, so we have $\left<v_1|v_2\right> = 0 = 0^\dagger = \left<v_2|v_1\right>$
$$\mu \left<v_1|v_2\right> = \left<v_1|A|v_2\right> = 0 $$
$$\lambda \left<v_2|v_1\right> = \left<v_2|A|v_1\right> = 0 = 0^\dagger = (\left<v_2|A|v_1\right>)^\dagger = \left<v_1\right|A^\dagger\left|v_2\right>$$
So we have $\left<v_1|A|v_2\right> = \left<v_1\right|A^\dagger\left|v_2\right>$. I don't think I can conclude that $A = A^\dagger$ that simply.
Am I missing something?
Thanks for your help.
EDIT
I was thinking of doing the following
$$ \left<v_1|A|v_2\right> \left|v_1\right> = \left< v_1 \right | A^\dagger \left | v_2 \right> \left| v_1 \right> $$
$$ \left<v_1|v_1\right> A\left|v_2\right> = \left< v_1 | v_1 \right> A^\dagger \left | v_2 \right> $$
Then the same for $\left|v_2\right>$ and get $||v_1||A||v_2|| = ||v_1||A^\dagger||v_2||$ and conclude that $A=A^\dagger$, but I'm not sure if I can reorder the bras and kets to draw this conclusion.
 A: Let $A$ be an $N\times N$ complex matrix $[a_{n,m}]$ for which there is an orthonormal basis of column vectors $[(e_{j})_{n}]$ which are eigenvectors of $A$ with corresponding eigenvalues $\lambda_{1},\lambda_{n},\cdots,\lambda_{N}$ where the $\lambda_{j}$ are real and some of the $\lambda_{j}$ may be repeated. Then
$$
           A\left[\begin{array}{c}(e_j)_1 \\ (e_j)_2 \\ \vdots \\ (e_j)_{N}\end{array}\right] = \lambda_{j} \left[\begin{array}{c}(e_j)_1 \\ (e_j)_2 \\ \vdots \\ (e_j)_{N}\end{array}\right]
$$
That gives
$$
           A\left[\begin{array}{cccc}
                    (e_1)_1 & (e_2)_1 & \cdots & (e_N)_1 \\
                    (e_1)_2 & (e_2)_2 & \cdots & (e_N)_2 \\
                    \vdots  & \vdots  & \ddots &  \vdots \\
                    (e_1)_N & (e_2)_N & \cdots & (e_N)_N
                  \end{array}\right]\\
        = \left[\begin{array}{cccc}
                    (e_1)_1 & (e_2)_1 & \cdots & (e_N)_1 \\
                    (e_1)_2 & (e_2)_2 & \cdots & (e_N)_2 \\
                    \vdots  & \vdots  & \ddots &  \vdots \\
                    (e_1)_N & (e_2)_N & \cdots & (e_N)_N
                  \end{array}\right]
          \left[\begin{array}{cccc}
                    \lambda_1 & 0 & \cdots & 0\\
                    0         & \lambda_2 & \cdots & 0 \\
                    \vdots  & \vdots  & \ddots &  \vdots \\
                    0 & 0 & \cdots & \lambda_N
                  \end{array}\right]
$$
The matrix $U$ formed from the orthonormal columns vectors is an orthogonal matrix, which guarantees that $U^{\star}U=UU^{\star}=I$ (I'm using $\star$ to denote conjugate transpose.) Therefore,
$$
                        AU = UD \implies A = UDU^{\star}.
$$
The diagonal matrix $D$ has only real eigenvalues. So $D^{\star}=D$. Now you get what you're trying to prove:
$$
    A^{\star}= (UDU^{\star})^{\star}=(U^{\star\star}D^{\star}U^{\star})=UDU^{\star}=A.
$$
A: If you assume that the set of orthogonal eigenvectors is complete, such a matrix is Hermitian.
It is easy to show this. If not, such a matrix may be not Hermitian.
