Matricial differential equation and eigenvalues

Question

I have the matricial differential equation:

$$\frac{\text{d}}{\text{d}t}\vec{x}(t)=A \cdot \vec{x}(t)$$

where:

$$\vec{x}(t) \in \mathbb{R}^3, \quad A=\begin{pmatrix} a & 0 & b \newline 0 & c & 0 \newline d & 0 & a \end{pmatrix}, \quad a,b,c,d \in \mathbb{R}$$

I need to find $a,b,c,d \in \mathbb{R}$ such that:

$$\lim_{t \to +\infty} \vec{x}(t)=\vec{0}$$

for every initial condition $\vec{x}(0) \in \mathbb{R}^3$.

The very short solution provided by the text is that the eigenvalues of the matrix $A$ must have negative real part. I can't undetstand why.

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What I have done so far:

I know that the solution can be written in the form:

$$\vec{x}(t)=e^{tA} \cdot \vec{x}(0)$$

and I can easily find the eigenvalues of $A$:

$$c, \quad a-\sqrt{bd}, \quad a+\sqrt{bd}$$

If the real part must be negative then we must have:

1. $c<0$;
2. $a<0 \quad \text{if} \quad bd<0$;
3. $a<-\sqrt{bd} \quad \text{if} \quad bd>0$.

Answer 1

If the matrix $A$ can be diagonalized then it can be written as

$$ A = U\Lambda U^{-1} $$

where $U$ is the matrix formed with the eigenvectors of $A$ and $\Lambda$ is the diagonal matrix formed with its eigenvalues.

$$ \Lambda = \pmatrix{\lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0 \\ 0 & 0 & \cdots & \lambda_n } $$

What's interesting about this representation is

\begin{eqnarray} A &=& U\Lambda U^{-1} \\ A^2 &=& (U\Lambda U^{-1})(U\Lambda U^{-1}) = U\Lambda^2 U^{-1}\\ &\vdots& \\ A^k &=& U\Lambda^k U^{-1} \end{eqnarray}

so that

$$ e^A = \sum_k \frac{A^k}{k!} = \sum_k\frac{U\Lambda^k U^{-1}}{k!} = U\left(\sum_k \frac{\Lambda^k}{k!}\right)U^{-1} = Ue^\Lambda U^{-1} $$

where

$$ e^\Lambda = \pmatrix{e^{\lambda_1} & 0 & \cdots & 0\\ 0 & e^{\lambda_2} & \cdots & 0 \\ 0 & 0 & \cdots & e^{\lambda_n} } $$

Now let's get back to your problem. As you correctly stated

$$ x(t) = e^{At}x(0) = U\pmatrix{e^{\lambda_1 t} & 0 & \cdots & 0\\ 0 & e^{\lambda_2t} & \cdots & 0 \\ 0 & 0 & \cdots & e^{\lambda_n t} }U^{-1}x(0) $$

from this last expression you can see that if you want the solution to go zero for every initial condition, then the term $e^{\lambda_1 t}, e^{\lambda_2 t}, \cdots$ must go to zero for large values of $t$, and that can only happen if the real part of all eigenvalues $\lambda_1,\cdots$ is negative