# Is there any hint how to prove this?

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

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

## 1 Answer

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 $$PD$$.