On a condition for skew-symmetry

Question

Let $A\in\mathbb{R}^{n\times n}$ be a generic lower triangular matrix and let $P\in\mathbb{R}^{n\times n}$ be a symmetric positive definite matrix.

True or false. Does $AP + PA^\top=0$ imply $AP=0$?

Answer 1

We can prove it by mathematical induction. The base case $n=1$ is trivial. For the inductive step, let $$ A=\pmatrix{L_1&0\\ M&L_2},\ P=\pmatrix{S_1&Y^T\\ Y&S_2}, \ AP=\pmatrix{L_1S_1&L_1Y\\ MS_1+L_2Y&MY^T+L_2S_2}. $$ where $L_1$ and $S_1$ are non-empty square submatrices of the same sizes and so are $L_2$ and $S_2$.

From $AP+PA^T=0$, we get $L_1S_1+S_1L_1^T=0$. Therefore, by induction assumption, $L_1=0$. Hence the condition $AP+PA^T=0$ when evaluated at other subblocks amounts to \begin{align} &MS_1+L_2Y=0,\tag{1}\\ &MY^T+L_2S_2+(MY^T+L_2S_2)^T=0\tag{2} \end{align} Equation $(1)$ gives $M=-L_2YS_1^{-1}$. Substitute this into $(2)$, we get $$ L_2(S_2-YS_1^{-1}Y^T)+(S_2-YS_1^{-1}Y^T)L_2^T=0. $$ Since $S_2-YS_1^{-1}Y^T$ is the Schur complement of $S_1$ in $P$, it must be positive definite. Thus $L_2=0$ by induction assumption and $M=-L_2YS_1^{-1}=0$. Consequently, $A=0$.

Answer 2

We can prove a stronger statement:

If $A$ has a real spectrum, $P$ is positive definite and $K=AP$ is skew-Hermitian, then $A=0$.

Proof. $A = KP^{-1}$ is similar to the skew-Hermitian matrix $K_1=P^{-1/2}KP^{-1/2}$, so that $A$ and $K_1$ share the same spectrum. On one hand, the spectrum of $A$ by assumption is real. On the other hand, the spectrum of $K_1$ is imaginary because it is skew-Hermitian. Thus $A$ and $K_1$ are nilpotent. Since the only nilpotent skew-Hermitian matrix is zero, $K_1=0$ and in turn $A=0$.