# If $T$ is a positive operator, and with $S^2=T$, then $S$ must be self-adjoint.

Let $$S$$ and $$T$$ be a linear operator on finite dimensional vector space $$V$$. If $$T$$ is a positive operator, and with $$S^2=T$$, then $$S$$ must be self-adjoint. Does this statement hold?

I try to use the polar decomposition of operators, there exists $$U$$ unitary, $$R$$ positive so that $$T=UR$$.

I found a similar question: Show there exists $S$ which is self-adjoint such that $S^2=T^*T$ and $S$ is invertible.

The statement is not true. Let $$S=\begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}$$ Then $$S^2=\begin{pmatrix} 1& 0\\ 0 &1\end{pmatrix}$$ but $$S$$ is not self-adjoint.
• Thanks. Can I ask how to get the adjoint of $S$? The definition is that $\langle Sv, v\rangle=\langle v, S^* v\rangle$ Mar 19, 2023 at 20:13
• The adjoint of a matrix $A$ is equal $\overline{A}^t,$ i.e. transposition plus complex conjugation of the matrix entries. When the entries are real numbers then $A^*=A^t.$ Mar 19, 2023 at 20:16
• Thank you! Can I ask why $S^2$ is positive? I know the definition of positive of operator but how about the matrix? Mar 19, 2023 at 20:34
• Positive $n\times n$ matrix satisfies by definition $\langle Ax,x\rangle \ge 0$ where $x\in \mathbb{C}^n$ and $\langle\cdot,\cdot \rangle$ is the standard inner product in $\mathbb{C}^n.$ The quantity $\langle Ax,x\rangle$ is equal $\sum_{k,l=1}^n a_{kl}x_l\overline{x_k}.$ The identity matrix is obviously positive as $\langle Ix,x\rangle=\sum_{k=1}^n|x|_k^2.$ I general a matrix is positive definite iff it is self-adjoint and its eigenvalues are nonnegative. Mar 19, 2023 at 20:40