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 $\left|v_1\right&gt;$ and $\left|v_2\right&gt;$ be two distinct orthogonal eigenvectors with distinct eigenvalues $\lambda$ and $\mu$ respectively for $A$, so we have $\left&lt;v_1|v_2\right&gt; = 0 = 0^\dagger = \left&lt;v_2|v_1\right&gt;$

$$\mu \left&lt;v_1|v_2\right&gt; = \left&lt;v_1|A|v_2\right&gt; = 0 $$
$$\lambda \left&lt;v_2|v_1\right&gt; = \left&lt;v_2|A|v_1\right&gt; = 0 = 0^\dagger = (\left&lt;v_2|A|v_1\right&gt;)^\dagger = \left&lt;v_1\right|A^\dagger\left|v_2\right&gt;$$

So we have $\left&lt;v_1|A|v_2\right&gt; = \left&lt;v_1\right|A^\dagger\left|v_2\right&gt;$. 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

$$ \left&lt;v_1|A|v_2\right&gt; \left|v_1\right&gt; = \left&lt; v_1 \right | A^\dagger \left | v_2 \right&gt; \left| v_1 \right&gt; $$
$$ \left&lt;v_1|v_1\right&gt; A\left|v_2\right&gt; = \left&lt; v_1 | v_1 \right&gt; A^\dagger \left | v_2 \right&gt; $$

Then the same for $\left|v_2\right&gt;$ 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.
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
$$
           A\left[\begin{array}{cccc}
                    (e_1)_1 &amp; (e_2)_1 &amp; \cdots &amp; (e_N)_1 \\
                    (e_1)_2 &amp; (e_2)_2 &amp; \cdots &amp; (e_N)_2 \\
                    \vdots  &amp; \vdots  &amp; \ddots &amp;  \vdots \\
                    (e_1)_N &amp; (e_2)_N &amp; \cdots &amp; (e_N)_N
                  \end{array}\right]\\
        = \left[\begin{array}{cccc}
                    (e_1)_1 &amp; (e_2)_1 &amp; \cdots &amp; (e_N)_1 \\
                    (e_1)_2 &amp; (e_2)_2 &amp; \cdots &amp; (e_N)_2 \\
                    \vdots  &amp; \vdots  &amp; \ddots &amp;  \vdots \\
                    (e_1)_N &amp; (e_2)_N &amp; \cdots &amp; (e_N)_N
                  \end{array}\right]
          \left[\begin{array}{cccc}
                    \lambda_1 &amp; 0 &amp; \cdots &amp; 0\\
                    0         &amp; \lambda_2 &amp; \cdots &amp; 0 \\
                    \vdots  &amp; \vdots  &amp; \ddots &amp;  \vdots \\
                    0 &amp; 0 &amp; \cdots &amp; \lambda_N
                  \end{array}\right]
$$
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:
$$
    A^{\star}= (UDU^{\star})^{\star}=(U^{\star\star}D^{\star}U^{\star})=UDU^{\star}=A.
$$
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.