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Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$ to obtain $$ \langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$.

Assume $\langle \psi_1\vert\psi_2\rangle \neq 0$ which means that $\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$. Since $\langle \psi_2\vert\psi_1\rangle \neq 0$, we may write $\vert\psi_2\rangle = \sqrt{\lambda}\vert\psi_1\rangle + \sqrt{(1-\lambda)}\vert\psi_1^\perp\rangle $ for some nonzero $\lambda$. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_1\vert M_2\vert\psi_1\rangle + \sqrt{(1-\lambda)}\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 \implies \langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 $$

Now you have a contradiction since it cannot be that both $\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0$ and $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$. $M_2\vert\psi_1\rangle = \vert\phi\rangle$, which you must be able to express in the basis $\{\vert\psi_1\rangle, \vert\psi_1^\perp\rangle\}$.

Hence, we conclude that $\langle\psi_1\vert\psi_2\rangle = 0$

Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$ to obtain $$ \langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$.

Assume $\langle \psi_1\vert\psi_2\rangle \neq 0$ which means that $\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$. Since $\langle \psi_1\vert\psi_2\rangle \neq 0$, we may write $\vert\psi_1\rangle = \sqrt{\lambda}\vert\psi_2\rangle + \sqrt{(1-\lambda)}\vert\psi_2^\perp\rangle $ for some nonzero $\lambda$. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_2\vert M_2\vert\psi_2\rangle + \sqrt{(1-\lambda)}\langle\psi_2\vert M_2\vert\psi_2^\perp\rangle = 0$$

But $\langle\psi_2\vert M_2\vert\psi_2\rangle = 1$, so we must have $\langle\psi_2\vert M_2 = \langle\psi_2\vert$ and hence, $\langle\psi_2\vert M_2\vert\psi_2^\perp\rangle = 0$. Therefore, $$\sqrt{\lambda}\langle\psi_2\vert M_2\vert\psi_2\rangle + \sqrt{(1-\lambda)}\langle\psi_2\vert M_2\vert\psi_2^\perp\rangle = \sqrt{\lambda} \neq 0 $$

We have arrived at a contradiction so we conclude that our assumption was wrong and $\langle\psi_1\vert\psi_2\rangle = 0$

Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$ to obtain $$ \langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$.

Assume $\langle \psi_1\vert\psi_2\rangle \neq 0$ which means that $\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$. Since $\langle \psi_2\vert\psi_1\rangle \neq 0$, we may write $\vert\psi_2\rangle = \sqrt{\lambda}\vert\psi_1\rangle + \sqrt{(1-\lambda)}\vert\psi_1^\perp\rangle $. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_1\vert M_2\vert\psi_1\rangle + \sqrt{(1-\lambda)}\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 \implies \langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 $$

Now you have a contradiction since it cannot be that both $\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0$ and $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$. $M_2\vert\psi_1\rangle = \vert\phi\rangle$, which you must be able to express in the basis $\{\vert\psi_1\rangle, \vert\psi_1^\perp\rangle\}$.

Hence, we conclude that $\langle\psi_1\vert\psi_2\rangle = 0$

Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$ to obtain $$ \langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$.

Assume $\langle \psi_1\vert\psi_2\rangle \neq 0$ which means that $\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$. Since $\langle \psi_2\vert\psi_1\rangle \neq 0$, we may write $\vert\psi_2\rangle = \sqrt{\lambda}\vert\psi_1\rangle + \sqrt{(1-\lambda)}\vert\psi_1^\perp\rangle $ for some nonzero $\lambda$. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_1\vert M_2\vert\psi_1\rangle + \sqrt{(1-\lambda)}\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 \implies \langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 $$

Now you have a contradiction since it cannot be that both $\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0$ and $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$. $M_2\vert\psi_1\rangle = \vert\phi\rangle$, which you must be able to express in the basis $\{\vert\psi_1\rangle, \vert\psi_1^\perp\rangle\}$.

Hence, we conclude that $\langle\psi_1\vert\psi_2\rangle = 0$

Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$ to obtain $$ \langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$.

Assume $\langle \psi_1\vert\psi_2\rangle \neq 0$ which means that $\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$. Since $\langle \psi_2\vert\psi_1\rangle \neq 0$, we may write $\vert\psi_2\rangle = \sqrt{\lambda}\vert\psi_1\rangle + \sqrt{(1-\lambda)}\vert\psi_1^\perp\rangle $. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_1\vert M_2\vert\psi_1\rangle + \sqrt{(1-\lambda)}\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 \implies \langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 $$

Now you have a contradiction since it cannot be that both $\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0$ and $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$. Hence, $\langle\psi_1\vert\psi_2\rangle = 0$

Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$ to obtain $$ \langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$.

Assume $\langle \psi_1\vert\psi_2\rangle \neq 0$ which means that $\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$. Since $\langle \psi_2\vert\psi_1\rangle \neq 0$, we may write $\vert\psi_2\rangle = \sqrt{\lambda}\vert\psi_1\rangle + \sqrt{(1-\lambda)}\vert\psi_1^\perp\rangle $. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_1\vert M_2\vert\psi_1\rangle + \sqrt{(1-\lambda)}\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 \implies \langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0 $$

Now you have a contradiction since it cannot be that both $\langle\psi_1^\perp\vert M_2\vert\psi_1\rangle = 0$ and $\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$. $M_2\vert\psi_1\rangle = \vert\phi\rangle$, which you must be able to express in the basis $\{\vert\psi_1\rangle, \vert\psi_1^\perp\rangle\}$.

Hence, we conclude that $\langle\psi_1\vert\psi_2\rangle = 0$