Suppose $V$ is a vector space with dimension $n$, I'm doubting that if operator $T^2$ is positive, then $T$ under some situations can be self-adjoint. However, I'm not certain about my reasoning.
My idea is that since $T^2$ is positive, then by the spectral theorem, there exists an orthonormal basis such that $T^2$ is diagonalized with all the entries on the diagonal are nonnegative, and they are also the eigenvalues of $T^2$.
Suppose $T^2$ has eigenvalues $\lambda_1,\cdots,\lambda_n$. Then each $\lambda_i$ should be the square of one eigenvalues of $T$. That is there should exist a number $\mu_i$ such that $\mu_i^2=\lambda_i$.
Also,$\lambda_i\geq 0$, so this implies that $\mu_i$ can't be complex. Since if $\mu_i=a+bi$ $$ (a+bi)^2=a^2+2abi-b^2 \geq 0$$ This implies that $a,b$ must one of them be zero, and if $a=0$ then $b=0$, and if $b=0$, $a$ can be any real numbers. Thus $b$ must be zero which implies that $\mu_i$ is real.
Then now if $T$ is normal and $\mathbb F=\mathbb C$(so $T$ is diagonalizable) and $T^2$ and $T$ are commutative, so under the same orthonormal basis, $T$ can be diagnosed with real numbers on the diagonal. Thus $T$ is self-adjoint.
If $T$ isn't normal, but $T^2$ has $n$ distinct eigenvalues. This implies that $T$ should have $n$ different real eigenvalues. Thus $T$ is diagonalizable. And since it is commutative with $T^2$, they can be diagonalized under the same orthonormal basis, which implies that $T$ is self-adjoint
Finally, If $T$ isn't normal, and $T^2$ doesn't have $n$ distinct eigenvalues. It's possible that the algebraic multiplicity of $\lambda_i$ for $T^2$ isn't the same as that for $\mu_i$. For example, on $\mathbb R^3$, $T^2$ has eigenvalues $5,5,5$ but $T$ only has one $\sqrt 5$, so $5$ and $\sqrt 5$ has different multiplicity. This results in that $T^2$ is diagonalizable, but $T$ can't, so $T$ shouldn't be self-adjoint.
I'm not sure whether I wrote above is correct. Any comments on this? Thanks!