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Let $A$ and $B$ be positive-definite matrices such that $A \le B.$ By matrix monotonicity of the root, this also implies that $A^{\alpha} \le B^{\alpha}$ for $\alpha \in [0,1].$ I am now curious under …

Let $f(z):=\langle g(z),g(z)\rangle,$ where $z \mapsto g(z)$ is holomorphic and $\langle \bullet,\bullet\rangle$ is an inner-product on some function space, such as $L^2$, such that $\langle g(z),g(z) …

This question is a more precise version of this question. Let's assume we have the matrix $$\left( \begin{array}{ccccc} 0 & a & 0 & 0 & 0 \\ f & 0 & b & 0 & 0 \\ 0 & e & 0 & c & 0 \\ 0 & 0 & d & 0 …

Consider the matrix $$A:=\left( \begin{array}{cccc} 0 & a & 0 & 0 \\ f & 0 & b & 0 \\ 0 & e & 0 & c \\ 0 & 0 & d & 0 \\ \end{array} \right)$$ I noticed that if I square this matrix then the eigenv …

Consider the matrix $$A_2:= \begin{pmatrix} a & b_1 \\ b_2 & a\end{pmatrix}.$$ Let $\sigma_2 = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$, then $$\sigma_2 A_2 \sigma_2 = \begin{pmatrix} a & -b_2 …