Why we don't write the derivative of the Lyapunov function as $\dot V = 2x^TPAx$

Question

Consider a LTI autonomous system like

$$ \dot x = Ax $$

and the Lyapunov function $V$ is defined as

$$ V(x) = x^T P x $$

where $P>0$. The derivative of $V$ with respect to time is

$$ \frac{d}{dt}V(x) = x^TA^TPx + x^TPAx = x^T \left( A^T P + P A \right) x $$

We have $\frac{d}{dt}V(x)$, a real number, and $x^TA^TPx$ , $x^TPAx$ are also two real numbers, and we know they are transpose to each other. Why don't we write the following instead?

$$\frac{d}{dt}V(x) = x^TPAx + \left(x^TPAx\right)^T = 2 x^T P A x $$

Then, if $PA<0$, we have the Lyapunov stability instead of $PA+A^TP < 0$. It seems maybe stupid but I can't figure it out. Thank you!

Note that what you differentiate with respect to time is $V \circ x$. — Rodrigo de Azevedo, Nov 13 '22 at 13:40
@RodrigodeAzevedo I tried if I calculate $\frac{d}{dt}(V) = (dV/dx)(dx/dt) $ and I get same resault. I must get into some trap but I can't figure it out.. — Zeze, Nov 13 '22 at 13:52
Note that only the symmetric part of $2 P A$ contributes to the quadratic form. What is this symmetric part? — Rodrigo de Azevedo, Nov 13 '22 at 13:53
@RodrigodeAzevedo $2PA$ is not symmetric but $x^TPAx$ is a number, so we have a transpose of this number is also itself which means $a^T = (x^TPAx)^T = x^TA^TPx = a^T = a = x^TPAx$ — Zeze, Nov 13 '22 at 13:54
Does it make sense to write $PA \prec 0$ when $PA$ is not symmetric? — Rodrigo de Azevedo, Nov 13 '22 at 13:58
@RodrigodeAzevedo Thank you so much! I got it. So in this logic$2PA$ should be symmetric to write $2PA < 0$ and if it is not, we take $(PA+A^TP)$ which is symmetric and it is exactly the usual case. — Zeze, Nov 13 '22 at 14:06

score 3 · Answer 1 · edited Apr 12 '23 at 16:21

3

Thank you @RodrigodeAzevedo ! The right logic should be like this:

Consider the expression $\frac{d}{dt}V(x) = x^T(A^TP + PA)x = 2x^TPAx$.

The objective is to find $P>0$ such that $\forall x\neq 0, \dot{V}(x) < 0$.

If the matrix $PA$ is symmetric, the condition $\dot{V}(x) < 0$ is equivalent to $PA < 0$, note that the notion of "negative/positive definite" is applicable only in case of symmetric matrix.
If not, the condition $\dot{V}(x) < 0$ is equivalent to $A^TP +PA < 0$ (where $A^TP +PA$ is a symmetric matrix).

So we always use the general form $A^TP +PA$ because it is already symmetric regardless of the symmetry of matrix $PA$.

edited Apr 12 '23 at 16:21

Rush Liu

3

answered Nov 13 '22 at 14:36

Zeze

161

What does this mean "note that the notion of `negative/positive definite' is applicable only in case of symmetric matrix."? – travelingbones Jan 16 '25 at 20:02
@travelingbones Take a look at this – Rodrigo de Azevedo Jan 16 '25 at 20:39

score 1 · Answer 2 · answered Jan 16 '25 at 20:11

Suppose $x^TMx \in \mathbb{R}$. This means $x^TMx = (x^TMx)^T = x^TM^tx$ (because the transpose of a 1x1 is itself.) Hence, $x^T(M-M^T)x = 0$.

Write $2M = (M+M^T) + (M-M^T)$. Note that in the quadratic form $$2x^tMx = x^T[(M+M^T) + (M-M^T)]x = x^T(M+M^T)x$$.

If you assume that $x^TPx>0$ and $P = P^T$ (it is not necessarily true that $x^TPx>0$ implies $P = P^T$ over $\mathbb{R}^n$, e.g., see example on Wikipedia of this), then $A^TP = (PA)^T$. This shows the equality of $^((A)^T+)=2^$ when $M = PA$.

Next note that $M+M^T$ is symmetric.

Why we don't write the derivative of the Lyapunov function as $\dot V = 2x^TPAx$

2 Answers2