線性矩陣不等式的常用引理：

Lemma 1：$M$ 是對稱陣，那么
$${\lambda }_{\max }(M)\le t?M?tI\le 0$$
Proof. Note that for an arbitrary matrix $M$ with an eigenvalue $s$ and a corresponding eigenvector $x,$ there holds
$$(M?tI)x=Mx?tx=(s?t)x$$
This states that for an arbitrary matrix $M$ there holds
$$\lambda (M?tI)=\lambda (M)?t$$
Thus, when $M$ is symmetric, we have
$$\begin{array}{rl}{\lambda }_{\max }(M)\le t& ?\lambda \max (M?tI)\le 0\\ & ?M?tI\le 0\end{array}$$

Lemma 2：$A$ 是具有合適維數(shù)的矩陣，$t$ 是一個正數(shù)，那么
$$A^{\mathrm{T}}A?t^{2}I\le 0?\left[\begin{array}{cc}?tI& A\\ A^{\mathrm{T}}& ?tI\end{array}\right]\le 0.$$
Proof: Put
$$Q=\left[\begin{array}{ll}I& A\\ 0& tI\end{array}\right]$$
then $Q$ is nonsingular since $t>0.$ Note that
$$Q^{\mathrm{T}}\left[\begin{array}{cc}?tI& A\\ A^{\mathrm{T}}& ?tI\end{array}\right]Q=\left[\begin{array}{cc}?tI& 0\\ 0& t\left(A^{\mathrm{T}}A?t^{2}I\right)\end{array}\right]$$
again in view of the fact that $t>0,$ it is clearly observed from the above relation that the equivalence in $(1.2.4)$ holds. Based on the above lemma, we can derive the following conclusion.

參考資料：

Matlab LMI(線性矩陣不等式)工具箱中文版介紹及使用教程

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