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Let's consider matrix M is defined as follows:

$M = \begin{bmatrix} P & v \\ v^t & d \end{bmatrix}$, where $P \succ 0$, d is a scalar, and v is a vector.

Problem: :To $M \succ 0$ be a positive definite matrix, the following inequality must be satisfied:

$d - v^tP^{-}v > 0$

Is there any hint how to prove this

Since M matrix has to be a positive definite matrix, $M = QDQ^t$, where some othonormal matrix Q and some diagonal matrix D. I am not sure this help to solve this or not? ALso I found something similar, Schur complement: https://en.wikipedia.org/wiki/Schur_complement

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  • $\begingroup$ Please use more descriptive titles. Also, what have you tried? $\endgroup$
    – Shaun
    Mar 31, 2022 at 20:34
  • $\begingroup$ @Shaun I added what I know, if you know could you give some hiht? if something not clear let me know $\endgroup$
    – GPrathap
    Mar 31, 2022 at 20:45
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    $\begingroup$ That's better. Please update the title though. I can't help you with the question as it's not really my area. $\endgroup$
    – Shaun
    Mar 31, 2022 at 20:52

1 Answer 1

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This is indeed a special case of Schur's complement. Hint: Recall that $M$ is $PD$ if and only if there exists an invertible matrix $N$ s.t $NMN^{T}$ is $PD$. Now, define $N$ to be $\begin{pmatrix}I & 0\\ -v^{T}P^{-1} & I \end{pmatrix}$, convince yourself that $N$ is invertible, and then show that indeed $NMN^{T}$ is $PD$ if and only if $d-v^{T}P^{-1}v$ is $PD$. Note that $P^{-1}$ exists since we know that $P$ is $PD$.

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