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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Any $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$) induces a semi-inner product on $F$ defined as: $$\langle\cdot,\cdot\rangle_{M}:F\times F\longrightarrow\mathbb{C},\;(x,y)\longmapsto\langle Mx, y\rangle,$$

Now let \begin{equation*} \mathbb{M}=\begin{pmatrix}A & B \\ C & D \end{pmatrix} \end{equation*} be a positive operator on $F\oplus F$. Does $\mathbb{M}$ induces a semi-inner product on $F\oplus F$?

Note that the inner product on $F\oplus F$ is defined as follows: If $x=\begin{pmatrix} x_1\\ x_2\end{pmatrix}\in F\oplus F$ with $x_1,x_2\in F$, and $x'=\begin{pmatrix}x_1'\\ x_2'\end{pmatrix}$ similarly, then $$\langle x,x'\rangle_{F\oplus F}:= \langle x_1,x_1'\rangle_F +\langle x_2,x_2'\rangle_F.$$

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As you say yourself, any positive bounded operator on a Hilbert space induces a semi-inner product. Your $\mathbb M$ is a positive bounded linear operator on the Hilbert space $F\oplus F$, so indeed it induces a semi-inner product on $F\oplus F$.

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  • $\begingroup$ How we define the semi inner product?I think M should be diagonal $\endgroup$ – Schüler Aug 23 '18 at 22:38
  • $\begingroup$ I fail to see what problem you perceive. You define $$(x,y)=\langle \mathbb M x,y\rangle. $$ $\endgroup$ – Martin Argerami Aug 24 '18 at 4:10

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