Asymptotic stability

What are eigenvalues and eigenvectors?

Consider a square matrix $F \in \mathbb{R}^{n \times n}$. If we can find a complex number $s \in \mathbb{C}$ and a non-zero, complex-valued vector $v \in \mathbb{C}^n$ that satisfy

\[(s I - F) v = 0\]

then we call $s$ an eigenvalue of $F$ and $v$ the corresponding eigenvector of $F$. If we wanted, we could rearrange this equation to put it in a form that might be more familiar:

\[0 = (s I - F) v = s v - F v \qquad\Rightarrow\qquad F v = s v.\]

One way to find eigenvalues is to solve the equation

\[\det (s I - F) = 0\]

where “$\det(\cdot)$” means taking a determinant. In general, this equation will be an $n$th-order polynomial in $s$, and so will have $n$ solutions — we might call them $s_1, \dotsc, s_n$. To find an eigenvector that corresponds to each eigenvalue $s_i$, we solve

\[F v_i = s_i v_i\]

for $v_i$. Note that there are many possible solutions to this equation and that eigenvectors are only unique up to a scale factor. In particular, for any real number $k \in \mathbb{R}$, we have

\[\begin{aligned} F (k v_i) &= k (F v_i) \\ &= k (s v_i) \\ &= s (k v_i). \end{aligned}\]

Apparently, if $v_i$ is an eigenvector corresponding to $s_i$, then so is $k v_i$ for any $k \neq 0$. For this reason, algorithms to find eigenvectors typically normalize them to have unit length.

How do I diagonalize a square matrix?

Suppose we have found the eigenvalues $s_1, \dotsc, s_n$ and eigenvectors $v_1, \dotsc, v_n$ of a square matrix $F\in\mathbb{R}^{n \times n}$. Define the matrix

\[V = \begin{bmatrix} v_1 & \dotsm & v_n \end{bmatrix}\]

with an eigenvector in each column, and also the matrix

\[\text{diag} (s_1, \dotsc, s_n) = \begin{bmatrix} s_1 & 0 & \dotsm & 0 \\ 0 & s_2 & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & s_n \end{bmatrix}\]

with the eigenvalues along the diagonal.

Two things are true.

First, the following equality holds:

\[F V = V \text{diag} (s_1, \dotsc, s_n)\]

You could easily verify this result for yourself.

Second, if $s_1, \dotsc, s_n$ are all distinct (i.e., if no two eigenvalues are the same), then the matrix $V$ is invertible. This result is harder to verify — it has to do with the fact that if the eigenvalues are distinct then the eigenvectors are linearly independent.

The key consequence of $V$ being invertible is that we can solve the above equality to write:

\[\text{diag} (s_1, \dotsc, s_n) = V^{-1} F V.\]

In this case — if all eigenvalues are distinct and so the matrix of eigenvectors is invertible — we say that $F$ is diagonalizable. The process of “diagonalizing $F$” is finding the matrix $V$.

What is the matrix exponential of a diagonal matrix?

It is easy to find the matrix exponential of a diagonal matrix, starting from the definition:

\[\begin{align*} e^{\text{diag} (s_1, \dotsc, s_n)t} &= I + \text{diag} (s_1, \dotsc, s_n)t + \frac{1}{2}\left( \text{diag} (s_1, \dotsc, s_n) t \right)^2 + \dotsm \\ &= \begin{bmatrix} 1 & 0 & \dotsm & 0 \\ 0 & 1 & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & 1 \end{bmatrix} + \begin{bmatrix} s_1t & 0 & \dotsm & 0 \\ 0 & s_2t & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & s_nt \end{bmatrix} + \begin{bmatrix} (s_1t)^2/2 & 0 & \dotsm & 0 \\ 0 & (s_2t)^2/2 & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & (s_nt)^2/2 \end{bmatrix} + \dotsm \\ &= \begin{bmatrix} 1 + s_1t + (s_1t)^2/2 + \dotsm & 0 & \dotsm & 0 \\ 0 & 1 + s_2t + (s_2t)^2/2 + \dotsm & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & 1 + s_nt + (s_nt)^2/2 + \dotsm \end{bmatrix} \\ &= \begin{bmatrix} e^{s_1t} & 0 & \dotsm & 0 \\ 0 & e^{s_2t} & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & e^{s_nt} \end{bmatrix} \end{align*}\]

What is the solution to a linear system that is diagonalizable?

We have seen that the solution to

\[\dot{x} = Fx\]

with the initial condition

\[x(0) = x_0\]

is

\[x(t) = e^{Ft}x_0.\]

Suppose $F$ is a diagonalizable matrix, so that

\[\text{diag} (s_1, \dotsc, s_n) = V^{-1} F V\]

where

\[\text{diag} (s_1, \dotsc, s_n)\]

is a diagonal matrix that contains the eigenvalues of $F$ and where

\[V = \begin{bmatrix} v_1 & \dotsm & v_n \end{bmatrix}\]

is a matrix of the corresponding eigenvectors. Then, applying the definition of matrix exponential again, we have

\[\begin{align*} e^{Ft}x_0 &= e^{V \text{diag} (s_1, \dotsc, s_n) V^{-1}t}x_0 \\ &= \left( I + V \text{diag} (s_1, \dotsc, s_n) V^{-1}t + \frac{1}{2}\left( V \text{diag} (s_1, \dotsc, s_n) V^{-1}t \right)^2 + \dotsm\right) x_0 \\ &= V \left( I + \text{diag} (s_1, \dotsc, s_n) t + \frac{1}{2} \left( \text{diag} (s_1, \dotsc, s_n)t \right)^2 + \dotsm\right) V^{-1}x_0 \\ &= V e^{\text{diag} (s_1, \dotsc, s_n)t}V^{-1}x_0 \\ &= V \begin{bmatrix} e^{s_1t} & 0 & \dotsm & 0 \\ 0 & e^{s_2t} & \dotsm & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dotsm & e^{s_nt} \end{bmatrix} V^{-1}x_0 \end{align*}\]

where the last step comes from what we just found out about the matrix exponential of a diagonal matrix. In this expression, the terms $V$, $V^{-1}$, and $x_0$ are constant. The only terms that depend on $t$, in fact, are the scalar exponential functions

\[e^{s_1t}, \dotsc, e^{s_nt}\]

that appear in the diagonal of

\[e^{\text{diag} (s_1, \dotsc, s_n)t}.\]

Therefore, we can infer the behavior of $x(t)$ based entirely on these scalar exponential functions. In particular, suppose that each eigenvalue $s_i$ — a complex number — has real part $a_i$ and imaginary part $b_i$, or in other words that

\[s_i = a_i + jb_i.\]

Euler’s formula tells us that

\[e^{s_it} = e^{(a_i+jb_i)t} = e^{a_it}\left(\cos(b_it) + j\sin(b_it)\right).\]

Apparently, as time gets large, one of three things is true about each of these terms:

  • if $a_i > 0$, then $e^{(a_i+jb_i)t}$ grows quickly
  • if $a_i = 0$, then $e^{(a_i+jb_i)t}$ is constant ($b_i=0$) or is sinusoidal with constant magnitude ($b_i \neq 0$)
  • if $a_i < 0$, then $e^{(a_i+jb_i)t}$ decays quickly to zero

It is possible to show that (more or less) the same result holds for any system $\dot{x}=Fx$, not only ones for which $F$ is diagonalizable. This takes more work, and involves the transformation of $F$ into Jordan normal form rather than into a diagonal matrix. We would discover that the terms that depend on $t$ all look like

\[t^me^{s_it}\]

where $m$ is an integer that is at most the multiplicity of the eigenvalue $s_i$. Since $e^{a_it}$ increases or decreases a lot faster than $t^m$, then the same three things we listed above would be true of each term in $x(t)$, just like before.

See the reference textbook for details.

When is a linear system asymptotically stable?

The system

\[\dot{x} = Fx\]

is called asymptotically stable if $x(t) \rightarrow 0$ as $t \rightarrow \infty$, starting from any initial condition $x(t_0) = x_0.$

Based on our observations about the solution to linear systems that are diagonalizable, we can state the following important result:

The system

\[\dot{x} = Fx\]

is asymptotically stable if and only if all eigenvalues of $F$ have negative real part.

In particular, we now have a test for whether or not a controller “works.” Suppose we apply linear state feedback

\[u = -Kx\]

to the state space system

\[\dot{x} = Ax + Bu\]

so that

\[\dot{x} = (A - BK)x.\]

The controller “works” when this system is asymptotically stable, i.e., when $x$ goes to zero as time gets large. We now know, therefore, that the controller “works” when all eigenvalues of $A - BK$ have negative real part.

We may not have a systematic way of finding a matrix $K$ to make the closed-loop system stable yet, but we certainly do have a systematic way now of deciding whether or not a given matrix $K$ makes the closed-loop system stable.