Are $A^TP+PA<0$, $P>0$ and $A^TP+PA\leq-I$, $P\geq I$ equivalent?

Question

Consider the LMI, where $A$ is a Hurwitz matrix:

$A^TP+PA<0$, $P>0$, minimize trace(P)

According to Stephen Boyd's book, the inequalities are homogeneous in $P$ and hence can by replaced with the nonstrict inequalities:

$A^TP+PA\leq-I$, $P\geq I$, minimize trace(P)

I do not understand why this is equivalent. Apparently the solution $P$ changes.

Answer 1

If $P$ is a solution to $A^\top P + P\,A \leq -I,\ P \geq I$ then scaling $P$ by some arbitrary small positive constant $\gamma$ the strict inequalities are also always satisfied, since

$$ \begin{array}{c} A^\top \gamma\,P + \gamma\,P\,A \leq -\gamma\,I < 0, \\ \gamma\,P \geq \gamma\,I > 0. \end{array} $$

So the solution to the nonstrict inequalities can always give you a solution to the strict inequalities.

Answer 2

Because $A^T P + P A$ and $P$ are homogenous functions of $P$, they are equivalent in the following sense:

1. Any solution of the non-strict inequalities ($A^T P + P A \leq -I$ and $P \geq I$) is necessarily a solution of the strict inequalities ($A^T P + P A < 0$ and $P > 0$).
2. For any solution of the strict inequalities there exists a solution of the non-strict inequalities (which itself will satisfy the non-strict inequalities)

We conclude that the LMI $A^T P + P A < 0$ and $P > 0$ has a solution if and only if $A^T P + P A \leq -I$ and $P \geq I$ has one.

See this answer for a detailed explanation of why homogeneity is needed to show this.

Answer 3

Here is a more formal proof that complements the other answers.

We want to solve the following feasibility problem $$ \begin{aligned} \textrm{find} \quad & X \\ \textrm{s.t.} \quad & X \succ 0 \\ \quad & A^T X + XA \prec 0 \end{aligned} \tag{1} $$ by solving the following feasibility problem instead $$ \begin{aligned} \textrm{find} \quad & X \\ \textrm{s.t.} \quad & X \succeq I \\ \quad & A^T X + XA \preceq -I \end{aligned} \tag{2} $$ There are two possible outcomes of the problem in $(2)$: Either an $X$ exists that satisfies the feasibility constraints, or it doesn't. When an $X$ exists that satisfies the feasibility constraints in $(2)$, we want to be able to say that this same $X$ also satisfies the feasibility constraints in $(1)$. Similarly, when an $X$ that satisfies the feasibility constraints in $(2)$ does not exist, then we want to be able to say that there also does not exist an $X$ that satisfies the feasibility constraints in $(1)$.

Therefore, we want to prove the following two statements:

1. If there exists an $X$ such that $X \succeq I$ and $A^T X + XA \preceq -I$, then $X \succ 0$ and $A^T X + XA \prec 0$.
2. If there does not exist a $Y$ such that $Y \succeq I$ and $A^T Y + YA \preceq -I$, then there does not exist an $X$ such that $X \succ 0$ and $A^T X + XA \prec 0$.

We first prove statement (1) as follows. Suppose that there exists an $X$ such that $X \succeq I$ and $A^T X + XA \preceq -I$. Because $X \succeq I$ and $I \succ 0$, then $X \succ 0$. Similarly, because $A^TX + XA \preceq -I$ and $-I \prec 0$, then $A^TX + XA \prec 0$, as desired. Note that we did not need to use the fact that $F(X) = A^TX + XA$ is homogeneous in $X$ here.

Next, we prove statement (2). We do so by proving the contrapositive of statement (2), which is: "If there exists an $X$ such that $X \succ 0$ and $A^TX + XA \prec 0$. then there exists a $Y$ such that $Y \succeq I$ and $A^TY + YA \preceq -I$". Suppose that there exists an $X$ such that $X \succ 0$ and $A^TX + XA \prec 0$. For any $X \succ 0$, there exists a $\lambda_1 \in (0,\lambda_\min(X)]$ such that $X - \lambda_1 I \succeq 0$, or $X \succeq \lambda_1 I$, where $\lambda_\min(X)$ is the smallest positive eigenvalue of $X$. Similarly, because $A^TX + XA \prec 0$, then there exists a $\lambda_2 \in (0,|\lambda_\max(A^TX + XA)|]$ such that $A^TX + XA + \lambda_2I \preceq 0$, or $A^TX + XA \preceq -\lambda_2I$, where $\lambda_\max(A^TX + XA)$ is the largest negative eigenvalue of $A^TX + XA$.

So, as a first step, choose $\lambda_1$ and $\lambda_2$ such that $\lambda_1 \leq \lambda_2$. This choice is guaranteed to exist, since when $\lambda_\min(X) > |\lambda_\max(A^TX + XA)|$, such that $\lambda_1 \in (0,\lambda_\min(X)]$ and $\lambda_2 \in (0,|\lambda_\max(A^TX + XA)|]$, we can choose $\lambda_1 \in (0,|\lambda_\max(A^TX + XA)| - \varepsilon]$ for a small enough $\varepsilon > 0$, and then choose $\lambda_2 \geq \lambda_1$. All other cases follow similarly.

Next, let $Y = \frac{1}{\lambda_1}X$. Since $X \succeq \lambda_1 I$ as shown above, then $Y \succeq I$, as desired. Similarly, since $A^TX + XA \preceq -\lambda_2I$, then we can multiply both sides of this inequality by $\frac{1}{\lambda_1}$ to get $$ \begin{align} \frac{1}{\lambda_1} \cdot \left(A^TX + XA\right) &\preceq -\frac{\lambda_2}{\lambda_1}I \\ A^T \left(\frac{1}{\lambda_1}X\right) + \left(\frac{1}{\lambda_1}X\right) A &\preceq -\frac{\lambda_2}{\lambda_1}I \\ A^T Y + Y A &\preceq -\frac{\lambda_2}{\lambda_1}I \end{align} $$ Because $\lambda_1 \leq \lambda_2$, then $-\frac{\lambda_2}{\lambda_1}I \preceq -I$, and so $A^T Y + YA \preceq -I$, as desired. Note that we used the fact that $F(X) = A^TX + XA$ is homogeneous in $X$ here.