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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.

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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.

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  • $\begingroup$ Thanks. Can I ask how to get the adjoint of $S$? The definition is that $\langle Sv, v\rangle=\langle v, S^* v\rangle$ $\endgroup$
    – Hermi
    Mar 19, 2023 at 20:13
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    $\begingroup$ 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 $A^*=A^t.$ $\endgroup$ Mar 19, 2023 at 20:16
  • $\begingroup$ Thank you! Can I ask why $S^2$ is positive? I know the definition of positive of operator but how about the matrix? $\endgroup$
    – Hermi
    Mar 19, 2023 at 20:34
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    $\begingroup$ 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. $\endgroup$ Mar 19, 2023 at 20:40

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